RBSE Solutions Class 11 Physics Chapter 8 Oscillatory Motion

RBSE Class 11 Physics Chapter 8 Textbook Exercises With Solutions

RBSE Class 11 Physics Chapter 8 Very Short Answer Type Questions

 

Question 1. For a simple harmonic motion what is the relationship between acceleration and displacement?
Answer: In simple harmonic motion, the acceleration of an object is directly proportional to its displacement from its central (mean) position. This acceleration always points back towards the mean position. This relationship is often represented by the formula \( F = - \omega^2 y \), where \( F \) is the restoring force, \( \omega \) is the angular frequency, and \( y \) is the displacement. The negative sign shows that the acceleration is always opposite to the direction of displacement.
In simple words: For simple harmonic motion, acceleration is always opposite to the direction of movement and gets bigger as the object moves further from its center.

🎯 Exam Tip: Remember that in SHM, acceleration and displacement are always in opposite directions, and acceleration is maximum at extreme positions, zero at the mean position.

 

Question 2. What will be the initial phase angle of a body starting its simple harmonic motion from the extreme end?
Answer: When a body starts its simple harmonic motion from the extreme end, it takes a time equal to one-fourth of its total time period (\( T/4 \)) to reach the mean position for the first time. Correspondingly, the initial phase angle for such a motion is \( \pi/2 \) radians. This phase angle helps describe the starting point of the oscillation.
In simple words: If an object starts its simple harmonic motion from the very edge, its starting phase angle will be \( \pi/2 \).

🎯 Exam Tip: The initial phase angle (or phase constant) determines the initial state of the oscillating particle at \( t=0 \), so carefully consider the starting point (mean position, extreme position, etc.).

 

Question 4. In simple harmonic motion which physical quantity is conserved?
Answer: In simple harmonic motion, the total mechanical energy of the system is conserved. This means that the sum of the kinetic energy (energy of motion) and potential energy (stored energy due to position) remains constant throughout the oscillation, assuming no external damping forces are present. This conservation is a key principle of ideal SHM.
In simple words: In simple harmonic motion, the total amount of energy (kinetic plus potential) always stays the same.

🎯 Exam Tip: The total mechanical energy is conserved only if there are no non-conservative forces like friction or air resistance acting on the system.

 

Question 5. What is the relation between oscillating frequency and total energy of a simple pendulum?
Answer: For a simple pendulum, the total energy (\( E_{total} \)) is related to its angular frequency (\( \omega \)) and amplitude (\( a \)) by the formula: \( E_{total} = \frac{1}{2} m\omega^2 a^2 \). Since angular frequency (\( \omega \)) is also related to linear frequency (\( n \)) by \( \omega = 2\pi n \), the total energy can also be written as \( E_{total} = \frac{1}{2} m(2\pi n)^2 a^2 = 2\pi^2 mn^2 a^2 \). This means the total energy is directly proportional to the square of the mass, the square of the frequency, and the square of the amplitude. Therefore, if the frequency increases, the total energy also increases significantly.
In simple words: The total energy of a pendulum depends on its mass, how far it swings (amplitude), and how fast it wiggles (frequency). More frequency means more total energy.

🎯 Exam Tip: Always remember that total energy in SHM is proportional to the square of both amplitude and angular frequency, making these factors very important.

 

Question 6. What will be the value of the total mechanical energy of a body of mass 0.1 kg having amplitude 3 cm and time period 2 s?
Answer:
Given values are:
Mass, \( m = 0.1 \text{ kg} \)
Amplitude, \( a = 3 \text{ cm} = 3 \times 10^{-2} \text{ m} = 0.03 \text{ m} \)
Time period, \( T = 2 \text{ s} \)
We need to find the total mechanical energy, \( E_{total} \).
The formula for total energy in simple harmonic motion is \( E_{total} = \frac{1}{2} m\omega^2 a^2 \).
First, we find the angular frequency \( \omega \) using the time period \( T \):
\( \omega = \frac{2\pi}{T} = \frac{2 \times 3.14}{2} = 3.14 \text{ rad/s} \)
Now, substitute the values into the total energy formula:
\( E_{total} = \frac{1}{2} m\omega^2 a^2 \)
\( E_{total} = \frac{1}{2} \times 0.1 \times (3.14)^2 \times (0.03)^2 \)
\( E_{total} = \frac{1}{2} \times 0.1 \times 9.8596 \times 0.0009 \)
\( E_{total} = 0.05 \times 9.8596 \times 0.0009 \)
\( E_{total} = 0.000443682 \)
\( E_{total} \approx 4.437 \times 10^{-4} \text{ J} \)
So, the total mechanical energy of the body is approximately \( 4.437 \times 10^{-4} \text{ J} \). This calculation demonstrates how mass, amplitude, and time period combine to determine the total energy.
In simple words: To find the total energy, we first calculate the angular speed using the time period. Then, we use the formula involving mass, angular speed squared, and amplitude squared to get the final energy.

🎯 Exam Tip: Always ensure all units are consistent (SI units are preferred) before plugging values into formulas to avoid errors in calculations.

 

Question 7. What will be the effect on time period on using soft spring of same length instead of hard spring?
Answer: The time period (\( T \)) of an oscillating body attached to a spring is given by the formula \( T = 2\pi \sqrt{\frac{m}{k}} \), where \( m \) is the mass and \( k \) is the spring constant. A softer spring has a smaller spring constant (\( k \)) compared to a harder spring. Therefore, if we replace a hard spring with a soft spring (meaning \( k_{soft} < k_{hard} \)), the value of \( \frac{m}{k} \) will increase. As a result, the time period of oscillation will increase, making the oscillation slower. A softer spring allows for a wider range of motion, thus extending the time for a complete oscillation.
In simple words: A soft spring has a small spring constant. Because time period is linked to the inverse of the spring constant, using a softer spring will make the oscillation take longer, meaning the time period increases.

🎯 Exam Tip: Remember that the time period of a spring-mass system is inversely proportional to the square root of the spring constant; a smaller \( k \) always means a larger \( T \).

 

Question 8. What will be the value of the time period of a simple pendulum if the amplitude is reduced to half of its original value?
Answer: The formula for the time period (\( T \)) of an ideal simple pendulum is given by \( T = 2\pi \sqrt{\frac{l}{g}} \), where \( l \) is the length of the pendulum and \( g \) is the acceleration due to gravity. This formula shows that the time period of a simple pendulum depends only on its length and the gravitational acceleration, not on its amplitude (for small oscillations). Therefore, if the amplitude of oscillation is reduced to half of its original value, the time period of the simple pendulum will remain unaffected. This is a crucial characteristic of simple harmonic motion.
In simple words: The time it takes for a simple pendulum to swing back and forth does not change even if you make it swing with a smaller amplitude.

🎯 Exam Tip: For small angular displacements (typically less than 10-15 degrees), the time period of a simple pendulum is considered independent of its amplitude.

 

Question 10. What will be the time period of a particle executing simple harmonic motion whose acceleration is \( a = -\frac{p}{q} x \)?
Answer:
Given the acceleration of the particle is \( a = -\frac{p}{q} x \).
For Simple Harmonic Motion (S.H.M.), the general equation for acceleration is \( a = -\omega^2 x \), where \( \omega \) is the angular frequency and \( x \) is the displacement.
Comparing the given equation with the standard S.H.M. equation, we can equate the coefficients of \( x \):
\( \omega^2 = \frac{p}{q} \)
Now, take the square root to find \( \omega \):
\( \implies \omega = \sqrt{\frac{p}{q}} \)
The time period (\( T \)) is related to the angular frequency (\( \omega \)) by the formula \( T = \frac{2\pi}{\omega} \).
Substitute the expression for \( \omega \):
\( \implies T = \frac{2\pi}{\sqrt{\frac{p}{q}}} \)
\( \implies T = 2\pi \sqrt{\frac{q}{p}} \)
So, the time period of the particle executing simple harmonic motion is \( 2\pi \sqrt{\frac{q}{p}} \). This formula helps calculate how long one full oscillation takes based on the parameters \( p \) and \( q \).
In simple words: By comparing the given acceleration with the standard formula, we find the angular frequency. Then, we use the angular frequency to calculate the time period for one full swing.

🎯 Exam Tip: Always remember the standard differential equation for SHM (\( a = -\omega^2 x \)) and its relation to the time period and angular frequency.

 

Question 11. The total energy of a simple pendulum is E. How much will be the kinetic energy and potential energy of a pendulum at an instant when the displacement is half the amplitude?
Answer:
The total energy (\( E \)) of a simple pendulum in Simple Harmonic Motion is given by \( E = \frac{1}{2} m\omega^2 a^2 \), where \( m \) is mass, \( \omega \) is angular frequency, and \( a \) is the amplitude.
The potential energy (\( U \)) of a simple pendulum at a displacement \( y \) is given by \( U = \frac{1}{2} m\omega^2 y^2 \).
We are interested in the instant when the displacement is half the amplitude, so \( y = \frac{a}{2} \).
Substitute \( y = \frac{a}{2} \) into the potential energy formula:
\( U' = \frac{1}{2} m\omega^2 \left(\frac{a}{2}\right)^2 \)
\( U' = \frac{1}{2} m\omega^2 \frac{a^2}{4} \)
Since \( E = \frac{1}{2} m\omega^2 a^2 \), we can write the potential energy in terms of total energy:
\( U' = \frac{E}{4} \)
Now, the kinetic energy (\( K \)) is the difference between the total energy and potential energy:
\( K' = E - U' \)
\( K' = E - \frac{E}{4} \)
\( K' = \frac{3E}{4} \)
Therefore, when the displacement is half the amplitude, the potential energy is \( \frac{E}{4} \) and the kinetic energy is \( \frac{3E}{4} \). This shows how energy is distributed between kinetic and potential forms at different points in the oscillation.
In simple words: When the pendulum is halfway from its center to its maximum swing, its stored energy is one-fourth of the total energy, and its motion energy is three-fourths of the total energy.

🎯 Exam Tip: Remember that kinetic energy is maximum at the mean position (\( y=0 \)) and zero at extreme positions (\( y=\pm a \)), while potential energy is the opposite.

RBSE Class 11 Physics Chapter 8 Short Answer Type Questions

 

Question 1. Prove that the projection of a uniform circular motion is simple harmonic motion.
Answer: Simple Harmonic Motion (SHM) is a special and basic type of oscillatory motion where a body moves back and forth along a straight line through its mean (center) position. The definition of SHM states that it is the projection of uniform circular motion onto the diameter of a circle. Imagine a particle of mass \( m \) moving with constant angular speed \( \omega_0 \) around the circumference of a circle with radius 'a' and center 'O' (as shown in some diagrams, e.g., fig. 8.2). At any moment, if we draw a perpendicular from the particle's position 'P' to the diameter (say, YY'), the point where it meets the diameter is 'M'. As the particle moves in a circle:

• When the particle is at point 'X' on the circle, the foot 'M' of the perpendicular is at 'O'.
• When the particle reaches point 'Y', 'M' also reaches 'Y'.
• When the particle reaches point 'X'', 'M' returns to 'O'.
• When the particle reaches point 'Y'', 'M' also reaches 'Y''.

This movement of point 'M' along the diameter, swinging back and forth from one end to the other through the center, perfectly describes simple harmonic motion. Thus, the projection of a particle in uniform circular motion onto any diameter is indeed simple harmonic motion. The continuous circular movement translates into a rhythmic linear oscillation.
In simple words: If an object moves in a perfect circle, its shadow on a straight line (like a wall next to the circle) will move back and forth in a simple harmonic motion.

🎯 Exam Tip: To prove this, it's essential to show that the projection's acceleration is directly proportional to its displacement from the mean position and always points towards the mean position.

 

Question 2. Write the characteristics and equation of S.H.M.
Answer: Simple Harmonic Motion (SHM) is a specific type of oscillatory motion defined by a restoring force that is directly proportional to the displacement and acts in the opposite direction. This relationship is often expressed as \( F = -ky \), where \( k \) is the spring constant and \( y \) is the displacement.

Characteristics of Simple Harmonic Motion:
(i) The motion always occurs along a straight line, moving back and forth (to and fro) around a specific central point, which is the equilibrium position.
(ii) The acceleration of the oscillating object is always directed towards the equilibrium state. This means it tries to bring the object back to its center.
(iii) In any position during oscillation, the acceleration of the particle is directly proportional to its displacement from the equilibrium position. The further it moves from the center, the stronger the acceleration pulling it back.

Equation of Simple Harmonic Motion:
Let's consider a particle oscillating to and fro around its equilibrium position. If the displacement of the particle at any instant is \( y \), then its acceleration is \( a = \frac{d^2 y}{dt^2} \).
According to Newton's second law, the force \( F \) acting on the particle is \( F = ma = m \frac{d^2 y}{dt^2} \).
For SHM, the restoring force is \( F = -ky \).
Equating the two expressions for force:
\( m \frac{d^2 y}{dt^2} = -ky \)
Rearranging the equation:
\( \implies m \frac{d^2 y}{dt^2} + ky = 0 \)
Divide by \( m \):
\( \implies \frac{d^2 y}{dt^2} + \frac{k}{m} y = 0 \)
We define \( \omega_0^2 = \frac{k}{m} \), where \( \omega_0 \) is the angular frequency.
So, the differential equation for linear simple harmonic motion is:
\( \implies \frac{d^2 y}{dt^2} + \omega_0^2 y = 0 \)
This equation mathematically describes the motion, showing that the acceleration is proportional and opposite to the displacement. The solution to this differential equation is typically of the form \( y = a \sin(\omega_0 t + \Phi) \), where \( a \) is the amplitude and \( \Phi \) is the initial phase angle.
In simple words: Simple harmonic motion happens when an object moves in a straight line, its acceleration always pulls it back to the middle, and this pull gets stronger the further it moves. The math equation for this motion connects its changing position with its speed and force.

🎯 Exam Tip: When writing the differential equation, clearly define each term and explain how it relates to the physical quantities of SHM.

 

Question 3. Graphically represent the displacement, velocity and acceleration for a time period of one cycle.
Answer:
(i) Displacement in SHM:
The displacement (\( y \)) of any particle in linear Simple Harmonic Motion at any instant \( t \) can be represented by the equation \( y = a \sin(\omega_0 t) \), where \( a \) is the amplitude and \( \omega_0 \) is the angular frequency. We can also write \( \omega_0 = \frac{2\pi}{T} \), so \( y = a \sin\left(\frac{2\pi}{T} t\right) \).
Let's see how \( y \) changes over one full time period \( T \):

This table shows how displacement changes, starting from zero, reaching maximum positive, returning to zero, reaching maximum negative, and finally returning to zero, completing one cycle. Graphically, this is a sinusoidal curve, starting at the origin, going up to \( +a \), down to \( -a \), and back to the origin, tracing a smooth wave. The displacement graph is a sine curve.

(ii) Velocity in SHM:
The velocity (\( v \)) is the derivative of displacement with respect to time: \( v = \frac{dy}{dt} = \frac{d}{dt} (a \sin(\omega_0 t)) = a\omega_0 \cos(\omega_0 t) \).
The velocity graph is a cosine curve, which starts at maximum positive, goes to zero, then maximum negative, and back to maximum positive. It is a quarter-cycle out of phase with the displacement.

(iii) Acceleration in SHM:
The acceleration (\( A \)) is the derivative of velocity with respect to time: \( A = \frac{dv}{dt} = \frac{d}{dt} (a\omega_0 \cos(\omega_0 t)) = -a\omega_0^2 \sin(\omega_0 t) \).
The acceleration graph is a negative sine curve, which starts at zero, goes to maximum negative, then zero, then maximum positive, and back to zero. It is half a cycle (\( \pi \) radians) out of phase with the displacement, meaning when displacement is positive, acceleration is negative, and vice-versa.
In simple words: The graph for displacement looks like a wave. The graph for velocity is also a wave, but it starts high when displacement is in the middle. The graph for acceleration is another wave, but it always goes the opposite way of displacement.

🎯 Exam Tip: When sketching these graphs, ensure the phase relationships are correct: velocity leads displacement by \( \pi/2 \), and acceleration leads displacement by \( \pi \) (or is \( \pi \) out of phase with it).

 

Question 3. Establish the expressions for potential energy, kinetic energy and total energy and prove conservation of the mechanical energy of a particle in S.H.M.
Answer:
(i) Kinetic Energy of a Simple Harmonic Oscillator:
The kinetic energy (\(K\)) of a particle is given by the formula:
\[ U_t = \frac{1}{2} ka^2 \sin^2 \omega_0 t \] Here, \( k = m\omega_0^2 \).
From the displacement equation, we know that \( \sin \omega_0 t = \frac{y}{a} \).
This means we can also write \( \cos^2 \omega_0 t = 1 - \sin^2 \omega_0 t = 1 - \frac{y^2}{a^2} \).
So, the kinetic energy of the particle is:
\[ K = \frac{1}{2} ka^2 \left(1 - \frac{y^2}{a^2}\right) \]
\( \implies K = \frac{1}{2} k(a^2 - y^2) \) ...(8.13)
This equation shows that kinetic energy is highest when \( y = 0 \) (at the mean position) and zero when \( y = \pm a \) (at the extreme positions). This is because the particle is moving fastest at the center and momentarily stops at the edges.
To find the average kinetic energy over one cycle, we use the formula:
\[ = \frac{1}{T} \int_0^T \frac{1}{2} ka^2 \cos^2 \omega_0 t \, dt \]
\( \implies = \frac{1}{2} ka^2 \frac{1}{T} \int_0^T \cos^2 \omega_0 t \, dt \)
We know that \( \int_0^T \cos^2 \omega_0 t \, dt = \frac{T}{2} \).
So, the average kinetic energy is:
\[ = \frac{1}{2} ka^2 \frac{T}{2T} \]
\( \implies = \frac{1}{4} ka^2 \)
(ii) Potential Energy of a Simple Harmonic Oscillator:
In simple harmonic motion, when a particle is moved from its equilibrium position, potential energy is stored in it. This energy comes from the work done against the restoring force \( F = -ky \). The increase in potential energy (\(U\)) is:
\[ U = \int_0^y (-F) dy = \int_0^y (ky) dy \]
\( \implies U = k \left[ \frac{y^2}{2} \right]_0^y \)
So, the potential energy is:
\[ U = \frac{1}{2} ky^2 \] ...(8.15)
If the displacement is given by \( y = a \sin \omega_0 t \), then the potential energy is:
\[ U_t = \frac{1}{2} ka^2 \sin^2 \omega_0 t \] ...(8.16)
Here, \( k = m\omega_0^2 \).
To find the average potential energy over one cycle, we use:
\[ = \frac{1}{T} \int_0^T U_t \, dt = \frac{1}{T} \int_0^T \frac{1}{2} m\omega_0^2 a^2 \sin^2 \omega_0 t \, dt \]
\( \implies = \frac{1}{2} ka^2 \frac{1}{T} \int_0^T \sin^2 \omega_0 t \, dt \)
We know that \( \int_0^T \sin^2 \omega_0 t \, dt = \frac{T}{2} \).
So, the average potential energy is:
\[ = \frac{1}{2} ka^2 \frac{T}{2T} \]
\( \implies = \frac{1}{4} ka^2 \) ...(8.17)
(iii) Total Energy of a Simple Harmonic Oscillator:
The total energy (\(E_{total}\)) of a simple harmonic oscillator is the sum of its instantaneous kinetic energy and instantaneous potential energy.
\[ E_{total} = K + U \]
\( \implies E_{total} = \frac{1}{2} k(a^2 - y^2) + \frac{1}{2} ky^2 \)

\( \implies E_{total} = \frac{1}{2} ka^2 - \frac{1}{2} ky^2 + \frac{1}{2} ky^2 \)

\( \implies E_{total} = \frac{1}{2} ka^2 \) ...(8.18)
This shows that the total energy of a particle in simple harmonic motion is directly proportional to the square of its amplitude (\(a\)) and the square of its angular frequency (\(\omega\)). Since \(k\) and \(a\) are constant, the total energy \(E_{total}\) remains constant throughout the motion. This confirms the law of conservation of mechanical energy for a simple harmonic oscillator, meaning energy is only converted between kinetic and potential forms. The total amount stays the same.
In simple words: The total energy in simple harmonic motion is always constant. It is the sum of kinetic and potential energy. As an object swings, its energy changes from potential (stored energy at high points) to kinetic (motion energy at the middle), but the total amount of energy never changes.

🎯 Exam Tip: Remember to clearly define kinetic and potential energy first, then derive their expressions and finally add them to show that the total energy is conserved and constant.

 

Question 4. Graphically represent the average kinetic energy per oscillation, potential energy per oscillation and also find out average kinetic energy and average potential energy per oscillation of simple harmonic motion.
Answer: The graph (Fig. 8.11, not provided in this output, but typically showing kinetic and potential energy curves) for a simple harmonic oscillator illustrates how kinetic energy (\(K\)), potential energy (\(U\)), and total energy (\(E_{total}\)) change with displacement (\(y\)). The displacement-potential energy curve is shaped like a parabola.
When studying such a graph, it becomes clear that during the motion, the sum of potential and kinetic energy is the total energy, and this total energy remains uniform. If at any displacement \(y\), the kinetic energy is equal to the potential energy, then:
\[ K = U \]
\( \implies \frac{1}{2} k(a^2 - y^2) = \frac{1}{2} ky^2 \)

\( \implies a^2 - y^2 = y^2 \)

\( \implies a^2 = 2y^2 \)

\( \implies y^2 = \frac{a^2}{2} \)

\( \implies y = \frac{a}{\sqrt{2}} \)
This means that at a displacement of \( y = \frac{a}{\sqrt{2}} \), the kinetic energy equals the potential energy. In simple harmonic motion, the total energy remains constant, but the kinetic and potential energies fluctuate, converting into each other. However, if external forces like friction are present, the total energy slowly decreases, causing the amplitude of oscillation to also decrease over time.
In simple words: In simple harmonic motion, kinetic energy and potential energy keep changing, but their sum (total energy) stays the same. The graph for potential energy looks like a U-shape. Kinetic and potential energy are equal when the object is about 70.7% of the way from the center to the edge.

🎯 Exam Tip: When drawing energy graphs for SHM, ensure the potential energy curve is parabolic, kinetic energy is an inverted parabola, and total energy is a straight horizontal line. Label key points where K=U, K=0, U=0, K=Max, U=Max.

 

Question 5. What is simple pendulum? For small displacement explain its motion. Calculate the formula for time period. On what quantities does it depend on?
Answer: An ideal simple pendulum is a theoretical concept with specific properties:
(i) Point mass: The bob (the mass at the end of the string) is assumed to be a single point, with its entire mass concentrated at its center. This is imaginary since real objects always have size.
(ii) Weightless string: The string connecting the bob to the suspension point is considered to have no mass.
(iii) Inextensible string: The string does not stretch or change its length when the pendulum swings.
(iv) Frictionless: There is no air resistance or friction at the suspension point.
In reality, a simple pendulum consists of a small, heavy metal ball (bob) suspended from a rigid support by a light, inextensible string. When the bob is pulled slightly to one side and released, it swings back and forth. For small displacements (angles less than about 10-15 degrees), the motion is approximately simple harmonic.
The restoring force that brings the pendulum back to its mean position is \( F = -mg \sin \theta \), where \(\theta\) is the angle of displacement. For small angles, \( \sin \theta \approx \theta \), so \( F \approx -mg \theta \). Since \( \theta = \frac{x}{L} \) (where \(x\) is the arc length displacement and \(L\) is the length of the string), we get \( F \approx -\frac{mg}{L} x \). This shows the force is proportional to displacement, meeting the condition for SHM.
The formula for the time period (\(T\)) of a simple pendulum is:
\[ T = 2\pi \sqrt{\frac{L}{g}} \] Here, \(L\) is the length of the pendulum (from the point of suspension to the center of gravity of the bob) and \(g\) is the acceleration due to gravity. From this formula, we can see that the time period of a simple pendulum depends on:
1. The length of the pendulum (\(L\)): A longer pendulum has a longer time period.
2. The acceleration due to gravity (\(g\)): A stronger gravitational field leads to a shorter time period.
The time period does NOT depend on the mass of the bob or the amplitude (for small oscillations). This makes pendulums useful for timekeeping.
In simple words: A simple pendulum is a weight swinging on a string. For small swings, it moves in a simple harmonic motion. How fast it swings (its time period) depends on the length of the string and gravity, but not on how heavy the weight is or how big the swing is (as long as it's a small swing).

🎯 Exam Tip: Clearly state the ideal conditions for a simple pendulum. Emphasize that for SHM, the displacement must be small, and remember the inverse relationship between time period and the square root of gravity.

 

Question 6. What is the difference between the series and parallel combination of springs? Calculate the value of effective spring constant for every combination.
Answer:
(a) Series Combination of Springs:
In a series combination, springs are connected end-to-end, one after another. When a mass is hung from this combination, the tension (force) applied to each spring is the same. However, the total extension is the sum of the extensions of individual springs.
Let's say we have two springs, \(S_1\) and \(S_2\), with spring constants \(k_1\) and \(k_2\), joined in series. If a force \(F\) is applied, causing total displacement \(y\), then:
For spring \(S_1\), the extension is \( y_1 = -\frac{F}{k_1} \).
For spring \(S_2\), the extension is \( y_2 = -\frac{F}{k_2} \).
The total displacement \(y\) is the sum of individual displacements: \( y = y_1 + y_2 \).
So, \( y = -\frac{F}{k_1} - \frac{F}{k_2} = -F \left( \frac{1}{k_1} + \frac{1}{k_2} \right) \).
Comparing this to \( F = -k_{eff} y \), where \(k_{eff}\) is the effective spring constant, we get:
\( -\frac{F}{k_{eff}} = -F \left( \frac{1}{k_1} + \frac{1}{k_2} \right) \)

\( \implies \frac{1}{k_{eff}} = \frac{1}{k_1} + \frac{1}{k_2} \) ...(8.29)
For \(n\) identical springs with constant \(k\), the effective spring constant would be \( \frac{1}{k_{eff}} = \frac{n}{k} \implies k_{eff} = \frac{k}{n} \).
The time period (\(T\)) for series combination is \( T = 2\pi \sqrt{\frac{m}{k_{eff}}} = 2\pi \sqrt{\frac{mn}{k}} \).
(b) Parallel Combination of Springs:
In a parallel combination, springs are connected side-by-side to a common base, and the mass is hung from their free ends. Here, the displacement (extension) of each spring is the same, but the total restoring force is the sum of the forces exerted by individual springs.
If we have two springs, \(S_1\) and \(S_2\), with spring constants \(k_1\) and \(k_2\), and they are extended by a displacement \(y\), then:
For spring \(S_1\), the restoring force is \( F_1 = -k_1 y \).
For spring \(S_2\), the restoring force is \( F_2 = -k_2 y \).
The total effective restoring force \(F\) on the combination is \( F = F_1 + F_2 \).
So, \( F = -k_1 y - k_2 y = -(k_1 + k_2) y \).
Comparing this to \( F = -k_{eff} y \), the effective spring constant is:
\[ k_{eff} = k_1 + k_2 \] For \(n\) identical springs with constant \(k\), the effective spring constant would be \( k_{eff} = nk \).
The time period (\(T\)) for parallel combination is \( T = 2\pi \sqrt{\frac{m}{k_{eff}}} = 2\pi \sqrt{\frac{m}{k_1 + k_2}} \). Springs connected in series become softer (larger extension), while springs in parallel become stiffer (smaller extension).
In simple words: When springs are in series (one after another), they stretch more easily, and the total stiffness is less. When they are in parallel (side-by-side), they become stiffer, and the total stiffness is more. Think of series as sharing the work, and parallel as combining strengths.

🎯 Exam Tip: Remember the analogy with resistors in circuits: springs in series combine like resistors in parallel, and springs in parallel combine like resistors in series for their constants. Always draw a simple diagram to visualize the force and displacement for each combination.

 

Question 7. Explain free, damped and forced oscillations with examples. Define resonance.
Answer:
1. Free Oscillations:
When an object vibrates at its natural frequency without any external force acting on it (apart from the initial displacement), it performs free oscillations. In an ideal scenario (no friction or air resistance), these oscillations would continue indefinitely with constant amplitude and energy. The frequency of these oscillations is called the natural frequency. An example is a simple pendulum swinging in a vacuum after being released, or a tuning fork vibrating after being struck (as long as external damping is ignored).
2. Damped Oscillations:
In reality, no system is perfectly ideal. Due to resistive forces like air resistance or friction, the energy of an oscillating system gradually decreases over time. This causes the amplitude of the oscillations to progressively reduce until the object comes to rest. These are called damped oscillations. A pendulum swinging in the air, slowly losing height with each swing, is a common example of damped oscillation. The energy is converted into heat or sound.
3. Forced Vibrations:
When an external periodic force is continuously applied to an oscillating system, forcing it to vibrate at the frequency of the external force, these are called forced vibrations. Even if the external force's frequency is different from the object's natural frequency, the object will vibrate at the driving force's frequency. For example, when a flour mill or a well motor runs, nearby objects like windows and doors often vibrate. This happens because the motor creates vibrations that force the other objects to vibrate. The external force keeps supplying energy to sustain the motion.
4. Resonance:
Resonance is a special case of forced vibration. It occurs when the frequency of the external periodic force applied to an oscillator becomes exactly equal to the natural frequency of the oscillator. In this condition, the amplitude of the forced oscillations becomes very large, often dramatically increasing. This is because maximum energy is transferred from the driving force to the oscillating system. A classic example is pushing a swing: if you push at the right time (at its natural frequency), the swing goes very high.
Examples of Resonance:
(i) When a vibrating tuning fork is placed on a hollow box containing air, if the natural frequency of the air in the box matches the tuning fork's frequency, a very loud sound is produced.
(ii) Soldiers are instructed not to march in step while crossing a bridge. If their marching frequency matched the bridge's natural frequency, resonance could cause large vibrations and potentially collapse the bridge.
(iii) During an earthquake, if the natural frequency of a building matches the frequency of the ground's vibrations, the building can vibrate violently and collapse.
(iv) Tuning a radio or television means adjusting its circuit's natural frequency to match the frequency of the desired broadcasting station, allowing strong signal reception.
(v) A train driver avoids blowing the whistle while crossing a bridge or tunnel to prevent the sound waves from causing resonance in the structure, which could cause damage.
In simple words: Free oscillations happen naturally without outside help. Damped oscillations slow down and stop because of friction. Forced vibrations happen when an outside push makes something shake. Resonance is a big shake that happens when the outside push matches the object's natural shaking speed, like a bridge breaking if soldiers march in step.

🎯 Exam Tip: Clearly distinguish between the causes and effects of each type of oscillation. For resonance, emphasize the critical condition where the driving frequency matches the natural frequency, leading to maximum amplitude. Provide at least one clear example for each concept.

 

RBSE Class 11 Physics Chapter 8 Numerical Questions

 

Question 1. A body executes simple harmonic motion according to the following equation; \( x = 5.0\cos(2\pi t + \frac{\pi}{4}) \) m where t = 1.5 s, calculate (i) displacement (ii) velocity (iii) acceleration.
Answer: Given the equation for simple harmonic motion: \( x = 5.0\cos(2\pi t + \frac{\pi}{4}) \)
We need to calculate (i) displacement, (ii) velocity, and (iii) acceleration at \( t = 1.5 \) s.
(i) Displacement (\(x\)):
Substitute \( t = 1.5 \) s into the displacement equation:
\( x = 5.0\cos(2\pi (1.5) + \frac{\pi}{4}) \)

\( \implies x = 5.0\cos(3\pi + \frac{\pi}{4}) \)
We know that \( \cos(3\pi + \theta) = -\cos(\theta) \). So,
\( x = 5.0\cos(\pi + \frac{\pi}{4} + 2\pi) = 5.0\cos(\pi + \frac{\pi}{4}) \)

\( \implies x = -5.0\cos(\frac{\pi}{4}) \)

\( \implies x = -5.0 \times \frac{1}{\sqrt{2}} \)

\( \implies x = -5.0 \times 0.7071 \)

\( \implies x = -3.5355 \) m
Therefore, the displacement is approximately \( -3.536 \) m.
(ii) Velocity (\(v\)):
From the given equation, the amplitude \(a = 5.0\) m and angular speed \( \omega = 2\pi \) rad/s. The general equation for velocity in SHM is \( v = -A\omega\sin(\omega t + \phi) \).
So, \( v = -5.0 \times 2\pi \sin(2\pi t + \frac{\pi}{4}) \)
At \( t = 1.5 \) s:
\( v = -10\pi \sin(2\pi (1.5) + \frac{\pi}{4}) \)

\( \implies v = -10\pi \sin(3\pi + \frac{\pi}{4}) \)
We know that \( \sin(3\pi + \theta) = -\sin(\theta) \). So,
\( v = -10\pi (-\sin(\frac{\pi}{4})) \)

\( \implies v = 10\pi \sin(\frac{\pi}{4}) \)

\( \implies v = 10\pi \times \frac{1}{\sqrt{2}} \)

\( \implies v = 10 \times 3.1416 \times 0.7071 \)

\( \implies v = 22.21 \) m/s
Therefore, the velocity is approximately \( 22.22 \) m/s.
(iii) Acceleration (\(f\)):
The general equation for acceleration in SHM is \( f = -\omega^2 x \).
We have \( \omega = 2\pi \) rad/s and \( x = -3.5355 \) m at \( t = 1.5 \) s.
\( f = -(2\pi)^2 \times (-3.5355) \)

\( \implies f = (4\pi^2) \times 3.5355 \)

\( \implies f = 4 \times (3.1416)^2 \times 3.5355 \)

\( \implies f = 4 \times 9.8696 \times 3.5355 \)

\( \implies f = 139.42 \) m/s\(^{-2}\)
Therefore, the acceleration is approximately \( 139.42 \) m/s\(^{-2}\). A positive acceleration indicates it's directed away from the mean position in the negative x direction.
In simple words: First, put the time into the equation to find how far the object is. Next, use the velocity formula with the time to find how fast it is moving. Finally, use the acceleration formula with the displacement to find how quickly its speed is changing. Remember that \( \pi/4 \) is 45 degrees, and \( 3\pi + \pi/4 \) is in the third quadrant.

🎯 Exam Tip: Pay close attention to the phase angle in trigonometric functions. Always use radians for calculations involving angular frequency. Remember the relationships: \( v = \frac{dx}{dt} \) and \( a = \frac{dv}{dt} = -\omega^2 x \).

 

Question 2. The time period of body executing simple harmonic motion is 2 s. After starting from t = 0 time after how much time its displacement will be equal to the amplitude?
Answer: Given: Time period \( T = 2 \) s.
The equation for displacement in simple harmonic motion, starting from \( t = 0 \) at the mean position, is \( y = a \sin(\omega t) \).
We know that \( \omega = \frac{2\pi}{T} \). So, \( \omega = \frac{2\pi}{2} = \pi \) rad/s.
Thus, the displacement equation becomes \( y = a \sin(\pi t) \).
We need to find the time \(t\) when the displacement \(y\) is equal to the amplitude \(a\).
Set \( y = a \):
\( a = a \sin(\pi t) \)

\( \implies 1 = \sin(\pi t) \)
The sine function equals 1 when its argument is \( \frac{\pi}{2}, \frac{5\pi}{2}, \ldots \). The first time this happens is at \( \frac{\pi}{2} \).
So, \( \pi t = \frac{\pi}{2} \)

\( \implies t = \frac{1}{2} \) s
Therefore, after \( 0.5 \) seconds, the displacement of the body will be equal to its amplitude. This happens when the body reaches an extreme end of its oscillation for the first time.
In simple words: We know how long one full swing takes (2 seconds). We want to find out when the object reaches its furthest point for the first time. Since it starts from the middle, it takes one-quarter of the total swing time to reach the very end. So, it takes 0.5 seconds.

🎯 Exam Tip: Remember that for SHM starting from the mean position, the object reaches its maximum displacement (amplitude) at \( t = \frac{T}{4} \), where \(T\) is the time period.

 

Question 3. A particle performing simple harmonic motion has maximum velocity 8 cm/s and has acceleration 16 cm/s² at a distance of 4 cm from the mean position. Calculate the time period and amplitude.
Answer: Given:
Maximum velocity, \( v_{max} = 8 \) cm/s
Acceleration, \( a_{dist} = 16 \) cm/s\(^2\), at displacement \( y = 4 \) cm.
We know the formula for maximum velocity in SHM: \( v_{max} = \omega a \) ...(1)
We also know the formula for acceleration at a given displacement \(y\): \( a_{dist} = -\omega^2 y \). Taking magnitude, \( a_{dist} = \omega^2 y \).
Substitute the given values into the acceleration formula:
\( 16 = \omega^2 \times 4 \)

\( \implies \omega^2 = \frac{16}{4} \)

\( \implies \omega^2 = 4 \)

\( \implies \omega = \sqrt{4} = 2 \) rad/s (angular frequency)
Now, use the value of \( \omega \) in the maximum velocity equation (1) to find the amplitude \(a\):
\( 8 = (2) \times a \)

\( \implies a = \frac{8}{2} \)

\( \implies a = 4 \) cm
So, the amplitude of the motion is \(4\) cm.
Finally, calculate the time period (\(T\)) using the angular frequency:
\( T = \frac{2\pi}{\omega} \)

\( \implies T = \frac{2\pi}{2} \)

\( \implies T = \pi \) s
Using the approximate value of \( \pi \approx 3.14 \), the time period is \(3.14\) s.
Therefore, the amplitude of the simple harmonic motion is \(4\) cm and the time period is \(3.14\) s. This shows how using key SHM formulas can help find unknown motion parameters.
In simple words: We are given the fastest speed and how much it slows down at a certain point. We use these to first find the angular speed (\(\omega\)). Once we have \(\omega\), we can figure out the maximum distance it travels (amplitude) and the time it takes for one full swing (time period).

🎯 Exam Tip: Remember the two key formulas for SHM: \( v_{max} = \omega a \) and \( a = \omega^2 y \). These are frequently used to solve problems involving unknown amplitude, angular frequency, or time period.

 

Question 5. A particle is doing simple harmonic motion. Its maximum velocity is 3 m/s and amplitude of motion is 0.2
Answer: This question appears incomplete as it states values but does not ask for a specific calculation. If the question intended to ask for angular frequency or time period, we could derive it. However, based on the verbatim text, we can only state the given information. For example, if we were asked for the angular frequency (\(\omega\)), we would use \( v_{max} = \omega a \).
Given:
Maximum velocity, \( v_{max} = 3 \) m/s
Amplitude, \( a = 0.2 \) m
From \( v_{max} = \omega a \), we could find \( \omega = \frac{v_{max}}{a} = \frac{3}{0.2} = 15 \) rad/s.
Then the time period \( T = \frac{2\pi}{\omega} = \frac{2\pi}{15} \approx 0.4189 \) s, which can be rounded to \( 0.42 \) s. This shows how knowing maximum velocity and amplitude allows calculation of other key motion properties.
In simple words: The fastest speed the object goes is 3 meters per second, and it swings 0.2 meters from the center. If we knew what to calculate, like how fast it spins (angular frequency) or how long one swing takes (time period), we could use these numbers.

🎯 Exam Tip: Even if a question seems incomplete, always identify the given parameters. Sometimes, the intent is simply to test your understanding of how these parameters relate to each other, even without an explicit "calculate" instruction.

 

Question 6. A piece of 1 kg mass is hanged from a spring. The spring constant is 50 N/m. The piece is stretched from the equilibrium position x = 0, t = 0 on a frictional surface to a distance x = 10 cm. The piece is at a distance 5 cm from its mean position. Calculate its kinetic, potential and total energy.
Answer: Given values:
Mass, \( m = 1 \) kg
Spring constant, \( k = 50 \) N/m
Amplitude, \( a = 10 \) cm \( = 0.1 \) m (This is the maximum stretch, so it's the amplitude)
Displacement from mean position, \( y = 5 \) cm \( = 0.05 \) m
First, calculate the angular speed (\(\omega\)):
\( \omega = \sqrt{\frac{k}{m}} \)

\( \implies \omega = \sqrt{\frac{50}{1}} = \sqrt{50} \approx 7.07 \) rad/s
Now, we can calculate the kinetic, potential, and total energy.
1. Kinetic Energy (\(K\)):
The formula for kinetic energy in SHM is \( K = \frac{1}{2} m\omega^2 (a^2 - y^2) \).
Substitute the values:
\( K = \frac{1}{2} \times 1 \times 50 ((0.1)^2 - (0.05)^2) \)

\( \implies K = 25 (0.01 - 0.0025) \)

\( \implies K = 25 (0.0075) \)

\( \implies K = 0.1875 \) J
Rounding to two decimal places, \( K \approx 0.19 \) J.
2. Potential Energy (\(U\)):
The formula for potential energy in SHM is \( U = \frac{1}{2} ky^2 \).
Substitute the values:
\( U = \frac{1}{2} \times 50 \times (0.05)^2 \)

\( \implies U = 25 \times 0.0025 \)

\( \implies U = 0.0625 \) J
3. Total Energy (\(E_{total}\)):
The total energy in SHM is \( E_{total} = \frac{1}{2} ka^2 \). This value remains constant throughout the motion.
Substitute the values:
\( E_{total} = \frac{1}{2} \times 50 \times (0.1)^2 \)

\( \implies E_{total} = 25 \times 0.01 \)

\( \implies E_{total} = 0.25 \) J
Alternatively, we can find total energy by adding kinetic and potential energies:
\( E_{total} = K + U = 0.1875 + 0.0625 = 0.25 \) J.
So, at 5 cm from the mean position, the kinetic energy is \(0.19\) J, the potential energy is \(0.0625\) J, and the total energy is \(0.25\) J. This problem shows how energy is distributed between kinetic and potential forms at different points in SHM.
In simple words: We have a spring with a weight. First, calculate the angular speed of the spring. Then, use this and other numbers to find the energy of movement (kinetic energy), the stored energy (potential energy) from the spring's stretch, and finally, add them to get the total energy. The total energy always stays the same.

🎯 Exam Tip: Ensure you use consistent units (meters for displacement, kilograms for mass). Remember that amplitude is the maximum displacement from the equilibrium position. The total energy can be calculated using either \( \frac{1}{2} ka^2 \) or \( K+U \).

 

Question 7. The time period of a simple harmonic oscillator is 6 s. After how much time will the displacement be half of its amplitude if it starts its motion from the equilibrium position?
Answer: Given:
Time period, \( T = 6 \) s
The equation for displacement of a particle starting its motion from the equilibrium position is:
\( y = a \sin(\omega t) \)
First, calculate the angular frequency (\(\omega\)):
\( \omega = \frac{2\pi}{T} = \frac{2\pi}{6} = \frac{\pi}{3} \) rad/s
We need to find the time (\(t\)) when the displacement (\(y\)) is half of its amplitude (\(a\)), i.e., \( y = \frac{a}{2} \).
Substitute this into the displacement equation:
\( \frac{a}{2} = a \sin(\frac{\pi}{3} t) \)

\( \implies \frac{1}{2} = \sin(\frac{\pi}{3} t) \)
We know that \( \sin(\frac{\pi}{6}) = \frac{1}{2} \). So,
\( \frac{\pi}{3} t = \frac{\pi}{6} \)

\( \implies t = \frac{\pi}{6} \times \frac{3}{\pi} \)

\( \implies t = \frac{1}{2} \) s
Therefore, after \( 0.5 \) seconds, the displacement will be half of its amplitude. This illustrates how the sinusoidal nature of SHM allows us to determine specific times for certain displacements.
In simple words: The object takes 6 seconds for one full swing. We want to find out when it is exactly halfway from the center to its furthest point. We calculate its spinning speed (\(\omega\)) and use the sine function to find that it takes half a second to reach that point.

🎯 Exam Tip: Always use the correct displacement equation: \( y = a \sin(\omega t) \) for starting from equilibrium, and \( y = a \cos(\omega t) \) for starting from an extreme position. Be careful with unit circle values for sine and cosine.

 

Question 8. The simple harmonic oscillator has total energy of 9.0 J at any instant with potential energy 5 J. The amplitude is 0.01 m and its mass is 2 kg. Calculate the time period.
Answer: Given values are: Total energy \( E_{\text{total}} = 9.0 \, \text{J} \), Amplitude \( a = 0.01 \, \text{m} \), Mass \( m = 2 \, \text{kg} \). We need to find the Time Period \( T \).
We know the formula for total energy in a simple harmonic oscillator:
\( E_{\text{total}} = \frac{1}{2} k a^2 \)
Substitute the given values:
\( 9.0 = \frac{1}{2} k (0.01)^2 \)
\( 9.0 = \frac{1}{2} k (0.0001) \)
To find \( k \), multiply both sides by 2 and divide by 0.0001:
\( 18 = k (0.0001) \)
\( k = \frac{18}{0.0001} \)
\( k = 180000 \, \text{N m}^{-1} \)
Now, we use the formula for the time period of a spring-mass system:
\( T = 2\pi \sqrt{\frac{m}{k}} \)
Substitute the values of \( m \) and \( k \):
\( T = 2\pi \sqrt{\frac{2}{180000}} \)
\( T = 2\pi \sqrt{\frac{1}{90000}} \)
\( T = 2\pi \frac{1}{300} \)
\( T = \frac{2 \times 3.14}{300} \)
\( T = \frac{6.28}{300} \)
\( T \approx 0.02093 \, \text{s} \)
Rounding off, we get:
\( T \approx 0.021 \, \text{s} \)
In simple words: We used the total energy formula to first find the spring constant, which tells us how stiff the spring is. Then, we used the spring constant and mass in another formula to figure out how long one full swing (oscillation) takes. The total energy in an ideal simple harmonic oscillator is the sum of its kinetic and potential energies, and this total energy remains constant if there's no friction or air resistance.

🎯 Exam Tip: Remember that total mechanical energy in SHM involves both the spring constant (k) and amplitude (a), while the time period is determined by mass (m) and spring constant (k).

 

Question 9. If the gravitational acceleration on Moon is 1/6th times the gravitational acceleration on the Earth. Then how many times will be the time period of the simple pendulum on moon?
Answer: Given that the gravitational acceleration on the Moon \( g_{\text{m}} \) is \( \frac{1}{6} \) times the gravitational acceleration on the Earth \( g_{\text{e}} \). So, \( g_{\text{m}} = \frac{g_{\text{e}}}{6} \). We need to find the time period on the Moon \( T_{\text{m}} \).
The formula for the time period of a simple pendulum is:
\( T = 2\pi \sqrt{\frac{L}{g}} \)
For Earth, the time period \( T_{\text{e}} \) is:
\( T_{\text{e}} = 2\pi \sqrt{\frac{L}{g_{\text{e}}}} \)
For the Moon, the time period \( T_{\text{m}} \) is:
\( T_{\text{m}} = 2\pi \sqrt{\frac{L}{g_{\text{m}}}} \)
Substitute \( g_{\text{m}} = \frac{g_{\text{e}}}{6} \) into the Moon's time period formula:
\( T_{\text{m}} = 2\pi \sqrt{\frac{L}{g_{\text{e}}/6}} \)
\( T_{\text{m}} = 2\pi \sqrt{\frac{6L}{g_{\text{e}}}} \)
We can rewrite this as:
\( T_{\text{m}} = 2\pi \sqrt{\frac{L}{g_{\text{e}}}} \times \sqrt{6} \)
Since \( T_{\text{e}} = 2\pi \sqrt{\frac{L}{g_{\text{e}}}} \), we can substitute \( T_{\text{e}} \):
\( T_{\text{m}} = T_{\text{e}} \sqrt{6} \)
So, the time period of a simple pendulum on the Moon will be \( \sqrt{6} \) times the time period on Earth. This means objects swing slower on the Moon due to weaker gravity.

🎯 Exam Tip: Remember that a simple pendulum's time period depends on the length of the pendulum and the local gravitational acceleration, but not on its mass or amplitude (for small oscillations).

 

Question 10. The bob of a simple pendulum has a mass of 0.025 kg. The effective length of the string is 1 m and time period 2.0 s. Calculate the value of gravitational acceleration g.
Answer: Given values are: Mass of the bob \( m = 0.025 \, \text{kg} \), Effective length of the string \( L = 1 \, \text{m} \), Time period \( T = 2.0 \, \text{s} \). We need to calculate the value of gravitational acceleration \( g \).
The formula for the time period of a simple pendulum is:
\( T = 2\pi \sqrt{\frac{L}{g}} \)
To find \( g \), we first square both sides of the equation:
\( T^2 = (2\pi)^2 \frac{L}{g} \)
\( T^2 = 4\pi^2 \frac{L}{g} \)
Now, rearrange the equation to solve for \( g \):
\( g = 4\pi^2 \frac{L}{T^2} \)
Substitute the given values:
\( g = 4 \times (3.14)^2 \times \frac{1}{(2.0)^2} \)
\( g = 4 \times 9.8596 \times \frac{1}{4} \)
\( g = 9.8596 \, \text{m s}^{-2} \)
Rounding to two decimal places, we get:
\( g \approx 9.86 \, \text{m s}^{-2} \)
In simple words: We used the formula for how long a pendulum takes to swing, then rearranged it to find the gravity. By plugging in the length of the string and the time it took to swing, we found the strength of gravity. This is a common way to measure gravity using a simple device.

🎯 Exam Tip: When solving for an unknown variable in a formula, always rearrange the equation first before substituting values to minimize calculation errors.

 

Question 11. The length of a second pendulum is 1 m [where g = 9.8m/s²]. What will be the length of the second pendulum on any planet where g = 4.9 m/s² ?
Answer: A second pendulum has a time period of \( T = 2 \, \text{s} \).
Given values: Length of second pendulum on Earth \( L_1 = 1 \, \text{m} \), Gravitational acceleration on Earth \( g_1 = 9.8 \, \text{m/s}^2 \).
We need to find the length of the second pendulum \( L_2 \) on a planet where gravitational acceleration \( g_2 = 4.9 \, \text{m/s}^2 \).
The formula for the time period of a simple pendulum is:
\( T = 2\pi \sqrt{\frac{L}{g}} \)
Since it's a second pendulum, \( T = 2 \, \text{s} \) on both Earth and the other planet.
So, for Earth:
\( 2 = 2\pi \sqrt{\frac{L_1}{g_1}} \)
And for the other planet:
\( 2 = 2\pi \sqrt{\frac{L_2}{g_2}} \)
Equating the two expressions for \( T \):
\( 2\pi \sqrt{\frac{L_1}{g_1}} = 2\pi \sqrt{\frac{L_2}{g_2}} \)
Square both sides and cancel \( 2\pi \):
\( \frac{L_1}{g_1} = \frac{L_2}{g_2} \)
Rearrange to solve for \( L_2 \):
\( L_2 = L_1 \frac{g_2}{g_1} \)
Substitute the given values:
\( L_2 = 1 \, \text{m} \times \frac{4.9 \, \text{m/s}^2}{9.8 \, \text{m/s}^2} \)
\( L_2 = 1 \times \frac{1}{2} \)
\( L_2 = 0.5 \, \text{m} \)
In simple words: A "second pendulum" always takes 2 seconds for one complete swing. Since gravity is weaker on the other planet, the pendulum needs to be shorter to swing at the same speed. We found that the length should be half of what it is on Earth. This shows how gravity affects the length needed for a pendulum to keep the same time.

🎯 Exam Tip: Understand that for a "second pendulum," the time period is constant (2s), allowing you to relate the length and gravitational acceleration directly between different locations.

 

Question 12. In simple harmonic motion in which position; velocity is half of the maximum velocity?
Answer: In simple harmonic motion (SHM), the velocity \( v \) at any displacement \( y \) is given by:
\( v = \omega_0 \sqrt{a^2 - y^2} \)
where \( \omega_0 \) is the angular frequency and \( a \) is the amplitude.
The maximum velocity \( v_{\text{max}} \) occurs at the mean position (\( y = 0 \)), so:
\( v_{\text{max}} = \omega_0 \sqrt{a^2 - 0^2} \)
\( v_{\text{max}} = \omega_0 a \)
We are looking for the position where the velocity \( v \) is half of the maximum velocity, i.e., \( v = \frac{v_{\text{max}}}{2} \).
Substitute \( v = \frac{\omega_0 a}{2} \) into the velocity equation:
\( \frac{\omega_0 a}{2} = \omega_0 \sqrt{a^2 - y^2} \)
Divide both sides by \( \omega_0 \):
\( \frac{a}{2} = \sqrt{a^2 - y^2} \)
Square both sides:
\( \left(\frac{a}{2}\right)^2 = a^2 - y^2 \)
\( \frac{a^2}{4} = a^2 - y^2 \)
Rearrange to solve for \( y^2 \):
\( y^2 = a^2 - \frac{a^2}{4} \)
\( y^2 = \frac{4a^2 - a^2}{4} \)
\( y^2 = \frac{3a^2}{4} \)
Take the square root of both sides to find \( y \):
\( y = \pm \sqrt{\frac{3a^2}{4}} \)
\( y = \pm \frac{a\sqrt{3}}{2} \)
So, the velocity is half of the maximum velocity when the particle is at a displacement of \( \frac{a\sqrt{3}}{2} \) from the mean position. This means the object is quite far from the center, slowing down as it approaches the end of its swing.

🎯 Exam Tip: Remember the key formulas for velocity in SHM at any displacement and maximum velocity. Setting up the ratio and solving algebraically is the standard approach for such problems.

 

Question 13. Calculate the percentage change in the time period of a simple pendulum in the following positions;
(i) If g is constant then T is proportional to \( \sqrt{L} \). If L is increased by 5%, then \( T_2 / T_1 = \sqrt{L_2 / L_1} \)
(ii) Even if mass is increased by 5% time period remains unaffected.
(iii) If amplitude is increased by 5%, the time period remains unaffected.
Answer: The time period of a simple pendulum is given by the formula:
\( T = 2\pi \sqrt{\frac{L}{g}} \)

(i) If gravitational acceleration (\( g \)) is constant, then the time period (\( T \)) is directly proportional to the square root of the length (\( L \)):
\( T \propto \sqrt{L} \)
If the length \( L \) is increased by 5%, then the new length \( L_2 \) will be \( L_1 + 0.05 L_1 = 1.05 L_1 \).
The ratio of the new time period (\( T_2 \)) to the original time period (\( T_1 \)) is:
\( \frac{T_2}{T_1} = \sqrt{\frac{L_2}{L_1}} \)
Substitute \( L_2 = 1.05 L_1 \):
\( \frac{T_2}{T_1} = \sqrt{\frac{1.05 L_1}{L_1}} \)
\( \frac{T_2}{T_1} = \sqrt{1.05} \)
\( \frac{T_2}{T_1} \approx 1.0247 \)
The percentage increase in time period is given by:
\( \text{Percentage increase} = \left( \frac{T_2 - T_1}{T_1} \right) \times 100\% \)
\( \text{Percentage increase} = \left( \frac{T_2}{T_1} - 1 \right) \times 100\% \)
\( \text{Percentage increase} = (1.0247 - 1) \times 100\% \)
\( \text{Percentage increase} = 0.0247 \times 100\% \)
\( \text{Percentage increase} \approx 2.47\% \)
Using the binomial approximation \( (1+x)^n \approx 1+nx \) for small \( x \):
\( \frac{T_2}{T_1} = (1.05)^{1/2} \approx 1 + \frac{1}{2} (0.05) = 1 + 0.025 = 1.025 \)
This means \( T_2 = 1.025 T_1 \).
So, the percentage increase in time period is approximately 2.5%.

(ii) The time period formula \( T = 2\pi \sqrt{\frac{L}{g}} \) does not include the mass (\( m \)) of the pendulum bob.
Therefore, if the mass is increased by 5%, the time period of the simple pendulum remains unaffected. It will continue to swing at the same rate.

(iii) For small oscillations, the time period formula \( T = 2\pi \sqrt{\frac{L}{g}} \) does not include the amplitude. This means the time it takes for one swing does not depend on how far you pull the pendulum back, as long as the angle is small.
Therefore, if the amplitude is increased by 5% (assuming it remains a small angle), the time period remains unaffected.
In simple words: The swing time of a simple pendulum mostly depends on its length and local gravity. Making the string longer increases the swing time, but changing the bob's weight or how wide it swings (if it's a small swing) does not change the time it takes. This is why pendulums are good for keeping time.

🎯 Exam Tip: Clearly state the formula for the time period of a simple pendulum and explain how each variable (length, gravity, mass, amplitude) affects it (or doesn't) to answer such questions comprehensively.

 

Question 14. A body of mass 0.5 kg hanged from an ideal spring is executing vertical oscillation. Oscillation time is \( \pi/2 \, \text{S} \), then calculate the spring constant.
Answer: Given values are: Mass of the body \( m = 0.5 \, \text{kg} \), Oscillation time (Time period) \( T = \frac{\pi}{2} \, \text{s} \). We need to calculate the spring constant \( k \).
The formula for the time period of a mass-spring system in vertical oscillation is:
\( T = 2\pi \sqrt{\frac{m}{k}} \)
Substitute the given values into the formula:
\( \frac{\pi}{2} = 2\pi \sqrt{\frac{0.5}{k}} \)
Divide both sides by \( \pi \):
\( \frac{1}{2} = 2 \sqrt{\frac{0.5}{k}} \)
Divide both sides by 2:
\( \frac{1}{4} = \sqrt{\frac{0.5}{k}} \)
Square both sides to remove the square root:
\( \left(\frac{1}{4}\right)^2 = \frac{0.5}{k} \)
\( \frac{1}{16} = \frac{0.5}{k} \)
Rearrange the equation to solve for \( k \):
\( k = 0.5 \times 16 \)
\( k = 8.0 \, \text{N m}^{-1} \)
The spring constant is \( 8.0 \, \text{N m}^{-1} \). This value tells us how stiff the spring is; a higher spring constant means a stiffer spring.

🎯 Exam Tip: Ensure you use the correct formula for the time period of a mass-spring system, \( T = 2\pi \sqrt{m/k} \), and correctly perform algebraic rearrangements to solve for the unknown variable.