GSEB Class 11 Physics Solutions Chapter 14 Oscillations

 

Question 1. Which of the following examples represent periodic motion?
1. A swimmer completing one (return) trip from one bank of a river to the other and back.
2. A freely suspended bar magnet displaced from its N-S direction and released.
3. A hydrogen molecule rotating about its centre of mass.
4. An arrow released from a bow.
Answer:
1. This is not a periodic motion. Although the swimmer's movement is back and forth, it will not have a fixed time period.
2. This is a periodic motion because if a freely hanging bar magnet is moved from its N-S direction and then let go, it swings around this position. Therefore, it also represents simple harmonic motion.
3. This is a periodic motion.
4. This is not a periodic motion.
In simple words: Periodic motion means something repeats regularly. A swimmer's trip back and forth might not take the same time each time. A magnet swinging, or a molecule spinning, will usually repeat in a steady way. An arrow, once shot, just moves forward, it does not repeat.

Exam Tip: When identifying periodic motion, look for movements that repeat in a regular cycle over a fixed time interval. Simple harmonic motion is a specific type of periodic motion with a restoring force proportional to displacement.

 

Question 2. Which of the following examples represent (nearly) simple harmonic motion and which represent periodic, but not simple harmonic motion?
(a) The rotation of Earth about its axis.
(b) Motion of an oscillating mercury column in a U-tube.
(c) Motion of a ball bearing inside a smooth curved bowl, when released from a point slightly above the lower most point.
(d) General vibrations of a polyatomic molecule about its equilibrium position.
Answer:
(a) This is periodic motion but not simple harmonic motion (S.H.M.) because it is not an oscillating movement around a fixed point.
(b) This is simple harmonic motion (S.H.M.).
(c) This is simple harmonic motion (S.H.M.).
(d) This is periodic motion but not simple harmonic motion (S.H.M.). A polyatomic molecule possesses many natural frequencies, and its overall movement results from combining several simple harmonic motions of various frequencies. This combination, or resultant movement, is periodic but not a simple harmonic motion.
In simple words: Simple harmonic motion (S.H.M.) is a special type of repeating movement that goes back and forth through a center point, like a pendulum. Earth's spin is regular but does not oscillate back and forth. A mercury column or a ball in a bowl will swing with S.H.M. A complex molecule might vibrate in a repeating way, but it is not a simple back-and-forth movement.

Exam Tip: Understand the difference between periodic motion (any motion that repeats regularly) and simple harmonic motion (a specific type of periodic motion where the restoring force is proportional to displacement).

 

Question 3. Given below are four x-t plots for linear motion. Which of the plots represent periodic motion? What is he period of motion (in case of periodic motion)?
Answer:
(a) This graph does not show periodic motion because the movement neither repeats itself nor returns to a central position.
(b) This graph illustrates periodic motion, with its time period being 2 seconds.
(c) This graph does not represent periodic motion because the movement does not repeat in an identical way.
(d) This graph shows periodic motion, and its time period is 2 seconds.
In simple words: Look at the graphs. If a graph pattern keeps repeating exactly over a set time, it's periodic. The "period" is how long it takes for one full pattern to happen. Some graphs show motion that does not repeat or repeats differently each time.

Exam Tip: For x-t plots, periodic motion is indicated by a graph shape that repeats exactly over regular time intervals. The period is the length of one complete repeating cycle.

 

Question 4. Which of the following functions of time represent (i) Simple harmonic, (ii) periodic but not simple harmonic and (iii) non-periodic motion? Give period for each case of periodic motion: (t0 is any positive constant)
(a) Sin wt - coswt.
(b) Sin³3wt.
(c) 3cos(\(\frac{π}{4}\) – 2wt).
(d) cos wt + cos 3wt + cos 5wt.
(e) exp (-w²t²).
(f) 1 + wt + w²t².
Answer:
A function shows periodic motion if it repeats exactly after a fixed time interval.
It shows simple harmonic motion (S.H.M.) if it can be written solely in the form of a cosine or sine function like \( (\frac{2\pi}{T}t + \phi) \) where \( T \) is the time period.
(a) sin wt - cos wt \( = \sqrt{2}(\frac{1}{\sqrt{2}} \sin wt - \frac{1}{\sqrt{2}} \cos wt) \)
\( = \sqrt{2}(\sin wt \cos \frac{\pi}{4} - \cos wt \sin \frac{\pi}{4}) \)
\( = \sqrt{2} \sin(wt - \frac{\pi}{4}) \)
This represents simple harmonic motion with a period \( T = \frac{2\pi}{w} \) and a phase angle \( (-\frac{\pi}{4}) \) or \( \frac{7\pi}{4} \).
(b) \(\sin^2 wt = \frac{1}{4}(3 \sin wt - \sin 3wt)\)
(\(\sin 3A = 3 \sin A - 4 \sin^3 A\))
Here, each term \( \sin 3wt \) and \( \sin^3 wt \) separately represents S.H.M., but this is not the result of combining two S.H.Ms. Therefore, it will only represent periodic motion, but not simple harmonic motion. Its time period is \( = \frac{2\pi}{w} \).
(c) \(3 \cos(\frac{\pi}{4} - 2wt) = 3 \cos (2wt - \frac{\pi}{4})\)
(\(\cos(-\theta) = \cos \theta\))
This represents simple harmonic motion, and its time period is \( \frac{2\pi}{2\omega} = \frac{\pi}{\omega} \).
(d) cos wt + cos 3wt + cos 5wt
This represents periodic but not simple harmonic motion. Its time period is \( \frac{2\pi}{w} \). It can be observed that each term denotes a periodic function with a different angular frequency. Since the period is the smallest time interval after which a function repeats its value, cos wt has a period \( T = \frac{2\pi}{w} \), cos 3wt has a period \( \frac{2\pi}{3\omega} = \frac{T}{3} \), and cos 5wt has a period \( \frac{2\pi}{5w} = \frac{T}{5} \). The last two forms repeat after any whole number multiple of their period. Thus, each term in the sum repeats itself after \( T \), and therefore the total sum is a periodic function with a period \( \frac{2\pi}{w} \).
(e) exp (-w²t²):
This is an exponential function that steadily decreases as time increases and approaches zero as \( t \rightarrow\infty \). As a result, it never repeats itself. Therefore, it represents non-periodic motion.
(f) \(1 + wt + w^2t^2\). This also represents non-periodic motion (which is physically unacceptable because the function approaches infinity as \( t \rightarrow \infty \)).
In simple words: We check if the functions repeat in time and if they fit the simple up-and-down (sine/cosine) shape of S.H.M. If a function can be written as \( \sin(\text{something}) \) or \( \cos(\text{something}) \), it's S.H.M. If it repeats but is more complex, it's just periodic. If it never repeats, like a growing or shrinking value, it's not periodic.

Exam Tip: To identify S.H.M., look for functions that can be expressed as a single sine or cosine wave. For periodic motion, ensure the function repeats its values exactly over regular intervals. Non-periodic functions typically grow or decay without repetition.

 

Question 5. A particle is in linear simple harmonic motion between two points A and B 10 cm apart. Take the direction from A to B as the positive direction and give the signs of velocity, acceleration and force on the particle when it is
(a) at the end A.
(b) at the end B.
(c) at the mid point of AB going towards A.
(d) at 2 cm away from B going towards A.
(e) at 3 cm away from A going towards B, and
(f) at 4 cm away from B going towards A.
Answer:
Let the situation be depicted as per figure here.
AB = 10 cm (given), P = midpoint of AB.
Here, A and B show the two furthest points of an S.H.M. For speed, the direction from A to B is considered positive. For acceleration and force, the direction is considered positive if it points along AP and negative if it points along BP.
(a) At the far end A, the particle in S.H.M. briefly stops at its extreme position of motion. Thus, its speed is zero, acceleration is positive as it points along AP. Force is also positive as it points towards AP.
Answer: 0, +, +.
(b) At the far end B, speed is zero. Here, acceleration and force are negative as they point along BP, which is the negative direction.
Answer: 0, -, -.
(c) At the middle point of AB, moving towards A, the particle is at the mean position P, with a tendency to move along PA, meaning the negative direction. Hence, speed is negative, and both acceleration and force are zero.
Answer: -, 0, 0.
(d) At 2 cm from B, moving towards A, the particle is at point Q with a tendency to move along QP, which is the negative direction. Therefore, velocity, acceleration, and force are all negative.
Answer: -, -, -.
(e) At 3 cm from A, moving towards B. The particle is now at R with a tendency to move along RP, meaning the positive direction. So, velocity, acceleration, and force will all be positive.
Answer: +, +, +.
(f) At 4 cm from B, moving towards A. The particle is at T with a tendency to move along TP, which is the negative direction for speed. So, velocity is negative, and acceleration and force point along TP. Therefore, both are negative.
Answer: -, -, -.
In simple words: When a particle moves in S.H.M., its speed, acceleration, and force change signs depending on where it is and which way it is moving. At the ends (A and B), it stops for a moment, so speed is zero, but acceleration and force are highest, pulling it back to the middle. In the middle, speed is highest, but acceleration and force are zero. The direction of motion (positive or negative) changes as it moves back and forth.

Exam Tip: Remember that at the extreme positions of S.H.M., velocity is zero, and acceleration/force are maximum and directed towards the mean position. At the mean position, velocity is maximum, and acceleration/force are zero. Pay attention to the defined positive direction.

 

Question 6. Which of the following relationships between the acceleration a and the displacement x of a particle involve simple harmonic motion?
(a) a = 0.7x
(b) a = - 200x²
(c) a = - 10x
(d) a = 100x³
Answer:
A particle performs Simple Harmonic Motion (S.H.M.) if the acceleration (a) it produces meets these two criteria:
(i) The acceleration 'a' is directly proportional to the displacement (let's say y) from the central position, meaning \( a \propto y \).
(ii) The acceleration 'a' points towards the central position, which means it acts in the opposite direction to where y increases.
Mathematically, this is expressed as \( a = -\omega^2 y \), where \( \omega \) is the angular frequency.
(a) The relation \( a = 0.7x \) does not meet equation (1), so it does not show S.H.M.
(b) The relation \( a = -200x^2 \) does not meet equation (1), so it does not show S.H.M.
(c) The relation \( a = -10x \), where x is the displacement, meets equation (1), so it represents S.H.M.
(d) The relation \( a = 100x^3 \) again does not meet equation (1), so it does not represent S.H.M.
In simple words: For simple harmonic motion, the pull (acceleration) on an object must always be proportional to how far it is from the center, and always pull it back towards the center. So, the equation must look like \( a = -\text{constant} \times x \). Options (b) and (d) have \( x^2 \) or \( x^3 \), which is not simple. Option (a) has \( a = 0.7x \) which pulls it away, not back. Only (c) matches the rule.

Exam Tip: The defining characteristic of Simple Harmonic Motion is that the restoring force (and thus acceleration) is directly proportional to the displacement from equilibrium and acts in the opposite direction. Mathematically, this is \( a \propto -x \).

 

Question 7. The motion of a particle executing S.H.M. is described by the displacement function: \( x(t) = A \cos (wt + \phi) \).
If the initial (t = 0) position of the particle is 1 cm and its initial velocity is \( w \) cm/s, what are its amplitude and initial phase angle? The angular frequency of the particle is \( \pi s^{-1} \). If instead of the cosine function, we choose the sine function to describe the S.H.M.: \( x = B \sin (wt + \alpha) \), what are the amplitude and initial phase of the particle with the above initial conditions?
Answer:
(a) The displacement function is \( x(t) = A \cos (wt + \phi) \) (Equation 1).
At \( t = 0 \), the angular frequency \( \omega = \pi s^{-1} \), and the particle's position is \( x = 1 \text{ cm} \).
The initial velocity is \( v = \omega = \pi \text{ cm/s} \) (Equation 2).
From (1) and (2), we find:
\( 1 = A \cos (\pi \times 0 + \phi) \)
\( = A \cos \phi \)
Also, we know \( \omega = \frac{2\pi}{T} \), so this means \( T = \frac{2\pi}{w} = \frac{2\pi}{\pi} = 2s \).
If we differentiate (1) with respect to \( t \), we get:
\( \frac{d}{dt}(x) = -A \sin(wt + \phi) (\omega) \)
\( = -A\omega \sin (wt + \phi) \)
So, \( v = -A\omega \sin (wt + \phi) \) (Equation 4).
Using equations (2) and (4) again, we have:
\( \pi = -A \times \pi \times \sin(\omega \times 0 + \phi) \)
\( = -A\pi \sin \phi \) (Equation 5).
If we square (2) and (5) and then add them, we get:
\( 1^2 + 1^2 = A^2(\sin^2\phi + \cos^2\phi) = A^2 \)
This implies \( A = \sqrt{2} \text{ cm} \).
By dividing (5) by (3), we get:
\( \frac{1}{1} = -\frac{\sin\phi}{\cos\phi} = -\tan \phi \)
Thus, \( \tan \phi = -1 = -\tan \frac{\pi}{4} \)
\( = \tan (2\pi - \frac{\pi}{4}) \)
\( = \tan \frac{7\pi}{4} \)
Therefore, \( \phi = \frac{7}{4}\pi \).
(b) If we use \( x = B \sin (wt + \alpha) \)
Or \( x = B \cos [(wt + \alpha) - \frac{\pi}{2}] \) (Equation 6).
Also let \( v \) be the velocity in this situation.
This implies \( v' = \frac{d}{dt}(x) \)
\( = B\omega \cos (\omega t + \alpha - \frac{\pi}{2}) \)
So, \( \pi = -B \times \pi \sin (\pi \times 0 + \alpha - \frac{\pi}{2}) \)
\( = -B\pi \sin(\alpha - \frac{\pi}{2}) \)
This gives us \( 1 = -B \sin (\alpha - \frac{\pi}{2}) \) (Equation 8).
Squaring and adding equations (7) and (8) yields:
\( 1^2 + 1^2 = B^2[\sin^2(\alpha - \frac{\pi}{2}) + \cos^2(\alpha - \frac{\pi}{2})] \)
This simplifies to \( 2 = B^2 \).
Therefore, \( B = \sqrt{2} \text{ cm} \).
Dividing equation (7) by (8) gives: \( 1 = -\tan(\alpha - \frac{\pi}{2}) \)
So, \( \tan (\alpha - \frac{\pi}{2}) = -1 = -\tan \frac{\pi}{4} \)
\( = \tan (2\pi - \frac{\pi}{4}) = \tan \frac{7\pi}{4} \)
This implies \( \alpha - \frac{\pi}{2} = \frac{7\pi}{4} \)
Or, \( \alpha = \frac{7\pi+2\pi}{4} = \frac{9\pi}{4} \)
\( = 2\pi + \frac{\pi}{4} \)
Hence, \( \alpha = \frac{\pi}{4} \).
In simple words: For a particle moving in simple harmonic motion, we use its position and speed at the very start (time zero) to figure out how big its swing is (amplitude) and where it starts in its cycle (phase angle). We can use either a cosine or a sine wave equation, but the amplitude will be the same, and the phase angle will just shift by \( \frac{\pi}{2} \).

Exam Tip: When solving S.H.M. problems, always use the given initial conditions (position and velocity at \( t=0 \)) to determine the amplitude and phase angle. Remember trigonometric identities like \( \sin^2\theta + \cos^2\theta = 1 \) and \( \cos(\theta - \frac{\pi}{2}) = \sin\theta \).

 

Question 8. A spring balance has a scale that reads from 0 to 50 kg. The length of the scale is 20 cm. A body suspended from this balance, when displaced and released, oscillates with a period of 0.6s. What is the weight of the body?
Answer:
Given values:
Mass, \( m = 50 \text{ kg} \)
Greatest extension, \( y = 20 - 0 = 20 \text{ cm} = 0.2 \text{ m} \)
Time period, \( T = 0.6s \)
The maximum force, \( F = mg = 50 \times 9.8 = 490.0 \text{ N} \)
The spring constant, \( k = \frac{F}{y} = \frac{490}{0.2} \)
\( = \frac{490 \times 10}{2} = 2450 \text{ Nm}^{-1} \)
We know that the time period is given by the formula:
\( T = 2\pi \sqrt{\frac{m}{k}} \)
Or, we can calculate mass \( m \) using the formula: \( m = \frac{T^2 k}{4 \pi^2} = \frac{(0.6)^2 \times 2450}{4 \times 9.87} = 22.36 \text{ kg} \)
The weight of the body, \( W = mg = 22.36 \times 9.8 = 219.1 \text{ N} \)
\( = 22.36 \text{ kgf} \).
In simple words: First, we find the spring's stiffness (spring constant) using the maximum weight it can hold and how much it stretches. Then, we use the formula that connects the spring's stiffness, the mass, and the time it takes to complete one swing (period) to find the actual mass of the swinging body. Finally, we convert that mass to weight.

Exam Tip: For spring-mass systems, remember the relationship between time period, mass, and spring constant: \( T = 2\pi \sqrt{\frac{m}{k}} \). Use this to find unknown quantities. Ensure units are consistent (e.g., meters, kilograms, seconds).

 

Question 9. A spring having a spring constant \( 1200 \text{ Nm}^{-1} \) is mounted on a horizontal table as shown. A mass of \( 3.0 \text{ kg} \) is attached to the free end of the spring. The mass is then pulled sideways to a distance of \( 2.0 \text{ cm} \) and released. Determine
1. the frequency of oscillations.
2. the maximum acceleration of the mass, and
3. the maximum speed of the mass?
Answer:
Given values:
Spring constant, \( k = 1200 \text{ Nm}^{-1} \)
Mass, \( m = 3.0 \text{ kg} \)
Amplitude (maximum displacement), \( A = 2.0 \text{ cm} = 0.02 \text{ m} \)
1. We know the time period \( T = 2\pi\sqrt{\frac{m}{k}} \).
We also know that frequency, \( \nu = \frac{1}{T} \).
Therefore, \( \nu = \frac{1}{2\pi}\sqrt{\frac{k}{m}} = \frac{1}{2 \times 3.142} \times \sqrt{\frac{1200}{3}} \)
\( = \frac{20}{2 \times 3.142} = 3.18 \)
Hence, \( \nu = 3.18 s^{-1} \approx 3.2 s^{-1} \).
2. The acceleration is given by the formula \( a = -\omega^2 x = -\frac{k}{m} x \).
The maximum acceleration, \( |a_{max}| = \frac{k}{m} |X_{max}| \), where \( \omega = \sqrt{\frac{k}{m}} \).
This means acceleration will be greatest when displacement x is at its maximum.
Thus, \( a = \frac{1200}{3} \times 0.02 = 8.0 \text{ ms}^{-2} \).
3. The greatest speed of the mass is given by:
\( v = A\omega = A\sqrt{\frac{k}{m}} = 0.02 \times \sqrt{\frac{1200}{3}} \)
\( = 0.02 \times 20 = 0.40 \text{ ms}^{-1} \).
In simple words: For a spring with a mass, we first find its angular speed (how fast it rotates in an imaginary circle). Then, we can calculate how many times it swings per second (frequency). The biggest pull it feels (maximum acceleration) happens when it's furthest stretched. The fastest it moves (maximum speed) also relates to its biggest swing and angular speed.

Exam Tip: For S.H.M., remember that maximum acceleration is \( A\omega^2 \) and maximum velocity is \( A\omega \). Also, \( \omega = \sqrt{\frac{k}{m}} \) for a spring-mass system.

 

Question 10. In exercise 14.9 let us take the position of mass when the spring is unstretched as \( x = 0 \), and the direction from left to right as the positive direction of x - axis. Give x as a function of time t for the oscillating mass if at the moment we start the stopwater (t = 0), the mass is
(a) at the mean position.
(b) at the maximum stretched position.
(c) at the maximum compressed position.
In what way do these different functions for S.H.M. differ from each other in frequency, in amplitude or the initial phase?
Answer:
Since the mass's position is \( x = 0 \) when the spring is unstretched, the equation for displacement in the x-direction is:
\( x = A \sin \omega t \).
(The initial phase \( \phi = \frac{\pi}{2} \))
From Question 14.9, the amplitude \( A = 2 \text{ cm} = 0.02 \text{ m} \).
The angular frequency \( \omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{1200}{3}} = 20 \text{ s}^{-1} \).
(a) When the mass is at its mean position, from equation (1), we find:
\( x = 2 \sin 20t \) (Equation 2).
(b) When the spring is in its most stretched position, \( \phi = \frac{\pi}{2} \).
This implies \( x = A \sin (\omega t + \phi) \)
\( = 2 \sin (20t + \frac{\pi}{2}) \)
\( = 2 \cos 20t \) (Equation 3).
(c) At the most compressed position:
\( \phi = \frac{\pi}{2} + \frac{\pi}{2} = \pi \)
\( = A \sin (wt + \pi) \)
\( = -A \sin wt \)
Thus, \( x = A \cos wt = -2 \cos (20t) \) (Equation 4).
Looking at equations (2), (3), and (4), we observe that the functions differ only in their initial phase. Their amplitude (\( A = 2 \text{ cm} \)) and periods are the same, meaning \( T = \frac{2\pi}{w} = \frac{2\pi}{20} = \frac{\pi}{10} \text{ rad s}^{-1} \).
In simple words: When a spring-mass system starts from different points (middle, stretched, or compressed), the equation describing its motion will look slightly different. The main difference will be the "starting point" or initial phase, not how big the swing is (amplitude) or how fast it repeats (frequency).

Exam Tip: Different initial conditions in S.H.M. primarily affect the phase angle, not the amplitude or frequency, as these are determined by the system's physical properties.

 

Question 11. Figure below corresponds to two circular motions. The radius of the circle, the period of revolution, the initial position, and the sense of revolution (i.e. clock-wise or anticlock-wise) are indicated on each figure. Obtain the corresponding simple harmonic motion of the x-projection of the radius vector of the revolving particle P, in each case.
Answer:
(a) In this case, at \( t = 0 \), the line segment OP forms an angle of \( \frac{\pi}{2} \) with the x-axis. Since the motion is clockwise, the initial phase \( \phi = \frac{\pi}{2} \) radians. Therefore, the x-projection of OP at time t will give the equation of Simple Harmonic Motion as:
\( x = A \cos (\frac{2\pi}{T}t + \phi) \)
\( = 3 \cos (\frac{2\pi}{2}t + \frac{\pi}{2}) \)
(Because \( A = 3 \text{ cm, } T = 2s \))
Or, \( x = 3 \cos (\pi t + \frac{\pi}{2}) \)
Alternative method:
(a) At \( t = 0 \), \( A = -3 \text{ cm} \), \( T = 2s \), \( \phi = 0 \) and the motion is clockwise.
This means the equation of S.H.M. is:
\( x = A \sin (\frac{2\pi}{T}t + \phi) \)
\( = -3 \sin (\frac{2\pi}{2}t) \)
\( = -3 \sin (\pi t) \)
(b) Given \( A = 2\text{m} \), at \( t = 0 \), \( T = 4s \), \( \phi = \pi \), and the motion is anticlockwise.
This implies the equation of S.H.M is:
\( x = 2 \sin (\frac{2\pi}{T}t + \phi) \)
\( = 2 \sin (\frac{2\pi}{4}t + \pi) \)
\( = -2 \cos (\frac{\pi}{2}t) \text{ (x in metres)} \)
In simple words: To find the simple harmonic motion from a circular motion, we look at the 'shadow' of the particle on the x-axis. We use the circle's radius (amplitude), how long one full turn takes (period), and where the particle starts (initial phase) to write the x-position as a cosine or sine function over time.

Exam Tip: When deriving the equation for x-projection of circular motion, remember to account for the initial phase angle (\( \phi \)), the direction of rotation (clockwise or anticlockwise), and the period (T) to get the correct angular frequency (\( \omega = 2\pi/T \)).

 

Question 12. Plot the corresponding reference circle for each of the following simple harmonic motions. Indicate the initial (t = 0) position of the particle, the radius of the circle, and the angular speed of the rotating particle. For simplicity the snese of rotation may be fixed to be anticlock-wise in every case; (x is in cm and t is in s).
(a) \( x = -2 \sin (3t + \frac{\pi}{3}) \)
(b) \( x = \cos (\frac{\pi}{6} - t) \)
(c) \( x = 3 \sin (\pi t + \frac{\pi}{4}) \)
(d) \( x = 2 \cos \pi t \).
Answer:
(a) Given the equation: \( x = -2 \sin (3t + \frac{\pi}{3}) \)
We can rewrite this using trigonometric identities:
\( = 2 \cos (\frac{\pi}{2} + 3t + \frac{\pi}{3}) \)
\( = 2 \cos (3t + \frac{5\pi}{6}) \)
\( = 2 \cos (\frac{2\pi}{2\pi} 3t + \frac{5\pi}{6}) \) (Equation 1)
The reference circle for this motion is shown below. By comparing this equation with the standard form \( x = A \cos (\omega t + \phi) \), we find:
Amplitude, \( A = 2\text{cm} \).
Angular frequency, \( \omega = 3 \text{ rad/s} \).
Period, \( T = \frac{2\pi}{\omega} = \frac{2\pi}{3} \text{ s} \).
Initial phase, \( \phi = \frac{5\pi}{6} \).
The initial position of the particle (at \( t=0 \)) on the circle would be at an angle of \( \frac{5\pi}{6} \) from the positive x-axis. The angular speed of the rotating particle is \( \omega = 3 \text{ rad/s} \). The rotation is anticlockwise as specified.

(b) Given the equation: \( x = \cos (\frac{\pi}{6} - t) \)
We can rewrite this as:
\( = \cos (t - \frac{\pi}{6}) \)
\( = 1 \cos (\frac{2\pi}{2\pi} t - \frac{\pi}{6}) \) (Equation 2)
The reference circle for this motion is shown below. If we compare equation (2) with the standard form \( x = A \cos (\frac{2\pi}{T}t + \phi) \) (Equation 3):
We find:
Amplitude, \( A = 1 \text{ cm} \).
Angular frequency, \( \omega = 1 \text{ rad/s} \).
Period, \( T = 2\pi \text{ s} \).
Initial phase, \( \phi = -\frac{\pi}{6} \).
The initial position of the particle (at \( t=0 \)) on the circle would be at an angle of \( -\frac{\pi}{6} \) from the positive x-axis. The angular speed of the rotating particle is \( \omega = 1 \text{ rad/s} \). The rotation is anticlockwise as specified.

(c) Given the equation for part (c) is \( x = 3 \sin (\pi t + \frac{\pi}{4}) \).
We can rewrite this as:
\( x = 3 \cos (\pi t + \frac{\pi}{4} - \frac{\pi}{2}) \)
\( x = 3 \cos (\pi t - \frac{\pi}{4}) \)
The reference circle for this motion is shown below. By comparing this equation with the standard form \( x = A \cos (\frac{2\pi}{T}t + \phi) \):
We find:
Amplitude, \( A = 3 \text{ cm} \).
Angular frequency, \( \omega = \pi \text{ rad/s} \).
Period, \( T = \frac{2\pi}{\omega} = \frac{2\pi}{\pi} = 2 \text{ s} \).
Initial phase, \( \phi = -\frac{\pi}{4} \).
The initial position of the particle (at \( t=0 \)) on the circle would be at an angle of \( -\frac{\pi}{4} \) from the positive x-axis. The angular speed of the rotating particle is \( \omega = \pi \text{ rad/s} \). The rotation is anticlockwise as specified.

(d) Given the equation: \( x = 2 \cos \pi t \)
We can write this as:
\( x = 2 \cos (\frac{\pi}{1}t + 0) \) (Equation 5)
The corresponding reference circle for this motion is shown in Fig. (d) (not included in output as it is on page 15). By comparing equation (3) with (5), we find:
Amplitude, \( A = 2 \text{ cm} \).
Angular frequency, \( \omega = \pi \text{ rad/s} \).
Period, \( T = \frac{2\pi}{\omega} = \frac{2\pi}{\pi} = 2 \text{ s} \).
Initial phase, \( \phi = 0 \).
The initial position of the particle (at \( t=0 \)) on the circle would be at an angle of \( 0 \) from the positive x-axis. The angular speed of the rotating particle is \( \omega = \pi \text{ rad/s} \). The rotation is anticlockwise as specified.
In simple words: To draw the reference circle for each simple harmonic motion (S.H.M.), we first identify its amplitude (which is the radius of the circle), its angular speed, and its starting position (initial phase angle) from the given equation. We assume it rotates counter-clockwise.

Exam Tip: When plotting reference circles for S.H.M., correctly identify the amplitude as the radius, the angular frequency from the coefficient of 't', and the initial phase angle (\( \phi \)) from the equation. Ensure the initial position on the circle corresponds to \( \phi \) at \( t=0 \).

 

Question 13. Figure (a) shows a spring with force constant \( k \) clamped rigidly at one end and a mass \( m \) attached to its free end. A force \( F \) applied at the free end stretches the spring. Figure (b) shows the same spring with both ends free and attached to a mass \( m \) at either end. Each spring in figure (b) is stretched by the same force \( F \).
(a) What is the maximum extension of the spring in the two cases.
(b) If the mass in figure (a) and the two masses in figure (b) are released free, what is the period of oscillation in each case?
Answer:
Here, \( k \) represents the force constant of the spring.
\( m \) is the mass attached to the free ends.
\( F \) is the force applied at the free end.
Case (1):
(a) Let \( a \) be the acceleration created in the mass 'm' connected to the free end because of the applied force \( F \).
So, \( F = ma \) (i)
Let \( y_1 \) be the extension produced in the spring in Fig. (a).
\( F = -ky_1 \) (ii)
From (i) and (ii), we get:
\( -ky_1 = ma \)
\( \implies ma = m \frac{d^2y}{dt^2} \)
where \( a = \frac{d^2y}{dt^2} \).
So, \( \frac{d^2y}{dt^2} = -\frac{k}{m} y \)
where \( y \) is the displacement equal to \( y_1 \).
We also know that \( a = -\omega^2y \). (iv)
From (iii) and (iv), we get:
\( \omega^2 = \frac{k}{m} \)
or \( \omega = \sqrt{\frac{k}{m}} \)
Therefore, the maximum extension produced in the spring is \( y_1 = y \).
or \( y_1 = \frac{F}{k} \)
From equation (iv), we see that \( a = y \), and the mass performs S.H.M.
If \( T_1 \) is the time period of oscillation for mass \( m \), then:
\( T_1 = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{m}{k}} \) (by using (v))
Case (2):
(a) Let \( y_2 \) be the total extension produced in the spring when the two masses are released. As masses are equal, if \( y \) is the extension produced in the spring for each mass, then \( y_2 = y' + y' = 2y' \).
Also, as explained in case 1 (a), we have \( \frac{F}{k} = 2y' \).
or \( y' = \frac{1}{2} \frac{F}{k} \)
This means the displacement of each mass, \( y' = \frac{1}{2} \frac{F}{k} \).
So, \( y_2 = 2 \cdot \frac{F}{2k} = \frac{F}{k} \).
Thus, the acceleration produced in each mass is then:
\( \frac{d^2y'}{dt^2} = -\frac{F}{m} = -\frac{2ky'}{m} \)
But \( \frac{d^2y'}{dt^2} = -\omega^2y' \) in S.H.M.
So that \( \omega^2 = \frac{2k}{m} \).
or \( \omega = \sqrt{\frac{2k}{m}} \).
(b) If \( T_2 \) is the time period of oscillation for mass \( m \), then:
\( T_2 = \frac{2\pi}{\omega} = \frac{2\pi}{\sqrt{\frac{2k}{m}}} = 2\pi \sqrt{\frac{m}{2k}} \).
Aliter:
Let \( \mu \) be the reduced mass of the system for fig. (b) which represents the inertia factor.
\( \mu = \frac{m_1 m_2}{m_1 + m_2} \), here \( m_1 = m_2 = m \).
\( \mu = \frac{m \cdot m}{m + m} = \frac{m^2}{2m} = \frac{m}{2} \).
We know that \( T = 2\pi \sqrt{\frac{\text{inertia factor}}{\text{spring factor}}} \).
So, \( T = 2\pi \sqrt{\frac{\mu}{k}} = 2\pi \sqrt{\frac{m/2}{k}} = 2\pi \sqrt{\frac{m}{2k}} \).
Thus, the maximum extension of the spring in each case is \( \frac{F}{k} \).
(a) For case (a): \( T = 2\pi \sqrt{\frac{m}{k}} \).
(b) For case (b): \( T = 2\pi \sqrt{\frac{m}{2k}} \).
In simple words: In the first setup, a single spring holds one mass, so its extension and oscillation period depend on that single mass and spring. In the second setup, the spring is effectively split, and two masses are involved, which changes how much it extends and how fast it oscillates. We calculate the extension by force divided by the spring constant. The period of oscillation involves the mass and spring constant, but how they are connected (one end fixed or both ends moving) alters the formula.

Exam Tip: Remember to differentiate between the effective spring constant and the reduced mass when dealing with multiple springs or masses in oscillation systems. Understand how the setup affects the restoring force and the inertial mass for period calculations.

 

Question 14. The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of 1.0 m. If the piston moves with simple harmonic motion with an angular frequency of 200 rad/min, what is its maximum speed?
Answer:
Stroke of piston \( = \) 2 times the amplitude.
Let \( A \) be the amplitude. Stroke \( = 1 \) m (given).
So, \( 1 = 2A \).
or \( A = \frac{1}{2} \) m.
Angular frequency, \( \omega = 200 \) rad/min.
\( V_{max} = \) maximum speed \( = ? \)
We know that the maximum speed of the block when the amplitude is \( A \) is given by:
\( V_{max} = \omega A \)
\( V_{max} = 200 \times \frac{1}{2} = 100 \) m/min.
To convert to m/s: \( = \frac{100}{60} \) m/s\(^{-1} \).
\( = \frac{5}{3} \) m/s\(^{-1} \approx 1.67 \) m/s\(^{-1} \).
In simple words: The piston's full movement is its stroke, which is double its amplitude. Given the stroke, we find the amplitude. Then, using the angular frequency and amplitude, we can calculate the fastest speed the piston reaches during its simple harmonic motion.

Exam Tip: Always remember that the stroke of an oscillating object is twice its amplitude. Convert all units to SI (meters, seconds) before final calculation unless otherwise specified to avoid errors.

 

Question 15. What is the period of a simple pendulum on the surface of the moon if its time-period on the surface on Earth is 3.5 s? (g on the surface of Earth is 9.8 m s\(^{-2}\).)
Answer:
Let \( T_e \) be the time period on the surface of Earth \( = 3.5 \) s.
Let \( T_m \) be the time period on the surface of the Moon \( = ? \).
\( g_e \) = acceleration due to gravity on the surface of Earth \( = 9.8 \) m s\(^{-2} \).
\( g_m \) = acceleration due to gravity on the surface of the Moon \( = 1.7 \) m s\(^{-2} \).
Let \( L \) be the length of the simple pendulum.
Using the formula for time period of a simple pendulum:
For Earth: \( T_e = 2\pi \sqrt{\frac{L}{g_e}} \) (1)
For Moon: \( T_m = 2\pi \sqrt{\frac{L}{g_m}} \) (2)
Dividing (2) by (1), we get:
\( \frac{T_m}{T_e} = \frac{2\pi \sqrt{\frac{L}{g_m}}}{2\pi \sqrt{\frac{L}{g_e}}} = \sqrt{\frac{g_e}{g_m}} \)
So, \( T_m = T_e \sqrt{\frac{g_e}{g_m}} \)
\( T_m = 3.5 \times \sqrt{\frac{9.8}{1.7}} \)
\( T_m = 3.5 \times \sqrt{5.7647} \)
\( T_m = 3.5 \times 2.4 \)
\( T_m = 8.4 \) s.
In simple words: The time period of a pendulum depends on the gravity it's in. We know its time period on Earth and Earth's gravity. If we also know the moon's gravity, we can use a ratio of the formulas to find its time period on the moon. The moon's weaker gravity makes the pendulum swing slower, so its period is longer.

Exam Tip: Remember the formula for the period of a simple pendulum, \( T = 2\pi \sqrt{\frac{L}{g}} \). For comparison problems, often a ratio approach simplifies calculations by canceling out common terms like \( 2\pi \) and \( L \).

 

Question 16. Answer the following questions:
(a) Time period of a particle in S.H.M. depends on the force constant k and mass m of the particle: \( T = 2\pi \sqrt{\frac{m}{k}} \). A simple pendulum executes S.H.M. approximately. Why then is the time period of a pendulum independent of the mass of the pendulum?
(b) The motion of a simple pendulum is approximately simple harmonic for small angle oscillations. For larger angles of oscillation, of a more involved analysis shows that T is greater than. Think of a qualitative argument to appreciate this result.
(c) A man with a wrist watch on his hand falls from the top of a tower. Does the watch give the correct time during free fall?
(d) What is the frequency of oscillation of simple pendulum in a cabin that is freely falling under gravity?
Answer:
(a) For a simple pendulum,
\( F = -\frac{mg}{L}y = -ky \).
So, \( k = \frac{mg}{L} \).
Here \( k \) represents the spring factor or force constant, which is directly proportional to the mass \( m \) of the particle. The time period of a simple pendulum is given by:
\( T = 2\pi \sqrt{\frac{m}{k}} \)
Substituting the value of \( k \):
\( T = 2\pi \sqrt{\frac{m}{\frac{mg}{L}}} = 2\pi \sqrt{\frac{mL}{mg}} = 2\pi \sqrt{\frac{L}{g}} \).
Thus, \( m \) cancels out in both the denominator and numerator. That is why the time period of a simple pendulum does not depend on the mass of the bob.
(b) The effective restoring force that pulls the bob back to its mean position for a simple pendulum is \( F = -mg \sin \theta \).
When \( \theta \) is small, \( \sin \theta \approx \theta \), and the expression for the time period of the simple pendulum is given by \( T = 2\pi \sqrt{\frac{L}{g}} \).
For larger values of \( \theta \), we will take \( F = -mg \sin \theta \). This is definitely \( < mg\theta \) because \( \sin \theta < \theta \) for larger values of \( \theta \). Thus, the effective value of \( g \) becomes \( g \sin \theta \) for larger \( \theta \). So, clearly \( g \sin \theta < g \). As \( T \propto \frac{1}{\sqrt{g}} \), for larger values of \( \theta \), the effective value of \( g \) decreases, and hence \( T \) increases beyond \( T = 2\pi \sqrt{\frac{L}{g}} \).
(c) Yes, it gives correct time during free fall. The functioning of the wrist watch relies on spring action, meaning a spring-wound watch runs due to the potential energy stored in its spring. Acceleration due to gravity plays no role in how a watch operates. So, it shows the correct time.
(d) We know that the frequency of oscillation of a simple pendulum is given by \( \nu = \frac{1}{T} = \frac{1}{2\pi} \sqrt{\frac{g}{L}} \).
So, \( \nu \propto \sqrt{g} \).
As \( g = 0 \) (i.e., gravity disappears for a freely falling body), the frequency becomes zero.
In simple words: (a) A pendulum's period seems to depend on mass in its 'spring constant' part, but the mass also affects the restoring force, so they cancel out, making the period only about length and gravity. (b) For bigger swings, the restoring force is less effective, which slows down the pendulum, making its period longer than for small swings. (c) A wrist watch works using springs, not gravity, so it keeps correct time even when falling freely. (d) If there's no gravity, a simple pendulum won't swing at all, so its oscillation frequency is zero.

Exam Tip: When analyzing pendulums, remember that the "mass independence" of the period is an approximation valid for simple pendulums. For larger angles, S.H.M. assumptions break down, and gravity's role in a free-falling system differs from its role in a clock mechanism.

 

Question 17. A simple pendulum of length \( L \) and having a bob of mass \( M \) is suspended in a car. The car is moving on a circular track of radius \( R \) with a uniform speed \( v \). If the pendulum makes small oscillations in a radial direction about its equilibrium position, what will be its time period?
Answer:
Since the pendulum is oscillating in the radial direction, it is acted upon by an additional acceleration equal to the centripetal acceleration of the car on the circular track of radius \( R \). This is \( \frac{v^2}{R} \), where \( v \) is the uniform speed of the car. Thus, two accelerations are acting on the pendulum in mutually perpendicular directions:
(i) acceleration due to gravity (\( g \))
(ii) centripetal acceleration (\( a_c = \frac{v^2}{R} \))
If \( a \) is the resultant acceleration of the pendulum, then:
\( a = \sqrt{g^2 + a_c^2} \)
\( a = \sqrt{g^2 + (\frac{v^2}{R})^2} = \sqrt{g^2 + \frac{v^4}{R^2}} \).
If \( T \) is the time period of oscillation of the pendulum, then:
\( T = 2\pi \sqrt{\frac{L}{a}} \)
\( T = 2\pi \sqrt{\frac{L}{\sqrt{g^2 + \frac{v^4}{R^2}}}} \).
In simple words: When a car moves in a circle, it creates an extra acceleration outwards. For a pendulum inside, this extra pull combines with gravity. We find the combined, or resultant, acceleration by using the Pythagorean theorem since they act at right angles. Then, we use this new total acceleration in the standard pendulum period formula to find the new oscillation time.

Exam Tip: In non-inertial frames, always identify all effective forces or accelerations acting on the system. Here, the centripetal acceleration acts as an additional "effective gravity" component, altering the total effective gravitational acceleration, which directly impacts the pendulum's period.

 

Question 18. A cylindrical piece of cork with base area \( A \) and height \( h \) floats in a liquid of density \( \rho_1 \). The cork is depressed slightly and then released. Show that the cork oscillates up and down simple harmonically with a period \( T = 2\pi \sqrt{\frac{h\rho}{\rho_1 g}} \), where \( \rho \) is the density of the cork. (Ignore damping due to viscosity of the liquid.)
Answer:
Let \( X \) be the equilibrium position of a cylinder floating in a given liquid.
Let \( A \) = area of cross-section of the cylindrical piece of cork.
\( h \) = height of the cylindrical piece of cork.
\( \rho \) = density of the material of the cylindrical cork.
\( \rho_1 \) = density of the liquid in which it floats.
\( L \) = length of the cylindrical piece of cork dipping in the liquid up to point \( P \) in position \( X \).
\( W \) = weight of the cylindrical cork.
\( W_1 \) = weight of the liquid displaced by the cork.
\( V \) = its volume.
\( m \) = mass of cork \( = A h \rho \).
\( W = mg = (Ah\rho)g \).
So, \( W_1 \) = Area of cross-section of cork \( \times \) length of cylinder dipping in liquid \( \times \) density of liquid \( \times g = A L \rho_1 g \).
According to the law of flotation, \( W = W_1 \).
or \( Ah\rho g = A L \rho_1 g \).
or \( h = \frac{L \rho_1}{\rho} \) (1)
Let the cylinder be pushed into the liquid through a distance \( y = PQ \). Then the restoring force acting on the cylinder is given by:
\( F \) = - weight of the liquid displaced by portion \( PQ \) of the cylindrical cork.
\( F = -(Ay)\rho_1 g = -(A\rho_1 g)y \). (2)
or \( F = -ky \).
Here \( k = A\rho_1 g \) is the force constant.
If \( a \) is the acceleration produced in the cylindrical piece of cork, then:
\( a = \frac{F}{\text{mass of cork}} = \frac{-(A\rho_1 g)y}{Ah\rho} = -\frac{\rho_1 g}{\rho h} y \). (3)
Now, as the acceleration of the cylindrical cork is directly proportional to its displacement from the equilibrium position and acts towards the equilibrium position, the motion of the bob is simple harmonic with a time period \( T \) given by:
\( T = 2\pi \sqrt{\frac{\text{Displacement}}{\text{Acceleration}}} \).
\( T = 2\pi \sqrt{\frac{y}{\frac{\rho_1 g}{\rho h} y}} = 2\pi \sqrt{\frac{y \rho h}{y \rho_1 g}} = 2\pi \sqrt{\frac{h\rho}{\rho_1 g}} \).
Hence proved.
In simple words: When you push down a floating cork, it displaces more liquid, creating an upward force. This force tries to bring the cork back up, and because it's proportional to how much you pushed it down, the cork bobs up and down in a simple harmonic motion. The period of this motion depends on the cork's height and density, and the liquid's density and gravity.

Exam Tip: For problems involving buoyancy and S.H.M., remember that the restoring force is due to the change in buoyant force. The key is to show that this restoring force is directly proportional to the displacement and acts in the opposite direction. Be careful with density and volume terms.

 

Question 19. One end of a U-tube containing mercury is connected to a suction pump and the other end to the atmosphere. A small pressure difference is maintained between the two columns. Show that, when the suction pump is removed, the liquid column of mercury in the U-tube executes simple harmonic motion?
Answer:
The suction pump creates the pressure difference, causing mercury to rise in one limb of the U-tube. When the pump is removed, a net force acts on the liquid column due to the difference in mercury levels in the two limbs. This force causes the liquid column to execute Simple Harmonic Motion, which can be explained as follows:
Consider the mercury contained in a vertical U-tube up to levels \( P \) and \( Q \) in its two limbs.
Let \( \rho \) = density of the mercury.
Let \( L \) = total length of the mercury column in both the limbs.
Let \( A \) = internal cross-sectional area of the U-tube.
\( m \) = mass of mercury in U-tube \( = LA\rho \).
Let the mercury be depressed in the left limb to \( P \) by a small distance \( y \). Then it rises by the same amount in the right limb to position \( Q' \).
So, the difference in levels in the two limbs \( PQ' = 2y \).
Volume of mercury contained in the column of length \( 2y = A \times 2y \).
Therefore, \( m = A \times 2y \times \rho \).
If \( W \) is the weight of liquid contained in the column of length \( 2y \).
Then \( W = mg = A \times 2y \times \rho \times g \).
This weight generates the restoring force (\( F \)) which tends to bring the mercury back to its equilibrium position.
So, \( F = -2Ay\rho g = -(2A\rho g)y \).
If \( a \) is the acceleration produced in the liquid column, then:
\( a = \frac{F}{m} = \frac{-(2A\rho g)y}{LA\rho} = -\frac{2\rho gy}{L\rho} = -\frac{2gy}{L} \). (i)
Here \( h \) is the height of mercury in each limb.
From (i), it is clear that \( a \propto y \) and the negative sign indicates that it acts opposite to \( y \). So, the motion of mercury in the U-tube is simple harmonic in nature, having a time period (\( T \)) given by:
\( T = 2\pi \sqrt{\frac{\text{Displacement}}{\text{Acceleration}}} = 2\pi \sqrt{\frac{y}{\frac{2gy}{L}}} \).
\( T = 2\pi \sqrt{\frac{yL}{2gy}} = 2\pi \sqrt{\frac{L}{2g}} \).
In simple words: When the pressure difference making the mercury uneven is removed, the heavier side pushes down, and the lighter side rises, creating a restoring force. This force works to make the levels equal again. Because this restoring force is directly proportional to how much the levels are uneven, the mercury oscillates up and down, showing simple harmonic motion with a period depending on the length of the mercury column and gravity.

Exam Tip: To prove S.H.M., always demonstrate that the restoring force (or acceleration) is directly proportional to the displacement from equilibrium and directed towards the equilibrium position. For liquid columns, careful calculation of mass and restoring force is crucial.