Maharashtra Board Class 12 Physics Chapter 5 Oscillations Exercise Solutions

1. Choose the correct option.

(i) A particle performs linear S.H.M. starting from the mean position. Its amplitude is A and its time period is T. At the instance when its speed is half the maximum speed, its displacement x is
(a) \( \frac{\sqrt{3}}{2} A \)
(b) \( \frac{2}{\sqrt{3}} A \)
(c) \( A/2 \)
(d) \( \frac{1}{\sqrt{2}} A \)
Answer: (a) \( \frac{\sqrt{3}}{2} A \)
In simple words: This question asks for the position of a vibrating object when it has slowed down to half of its top speed. By using the math of oscillations, we find it happens when the object is about 86% of the way to its furthest point.

📝 Teacher's Note: This is a classic application of the velocity-displacement formula \( v = \omega\sqrt{A^2 - x^2} \). Remind students that maximum velocity occurs at the mean position and is equal to \( \omega A \).

🎯 Exam Tip: Memorize the relation between velocity and displacement. Squaring both sides of the equation early helps avoid calculation errors with square roots.

(ii) A body of mass 1 kg is performing linear S.H.M. Its displacement x (cm) at t (second) is given by \( x = 6 \sin (100t + \pi/4) \). Maximum kinetic energy of the body is
(a) 36 J
(b) 9 J
(c) 27 J
(d) 18 J
Answer: (d) 18 J
In simple words: We are calculating the most energy a 1kg moving object has. By looking at its speed and how far it swings, we use a simple energy formula to find that it reaches 18 Joules of energy at its fastest point.

📝 Teacher's Note: Ensure students convert displacement from cm to meters (\( 6 \text{ cm} = 0.06 \text{ m} \)) before calculating energy. This is the most common place where marks are lost.

🎯 Exam Tip: Maximum kinetic energy is equal to total energy in SHM, given by \( \frac{1}{2} m\omega^2 A^2 \). Always check your units (SI units) before plugging numbers into the formula.

(iii) The length of the second’s pendulum on the surface of the earth is nearly 1 m. Its length on the surface of the moon should be [Given: acceleration due to gravity (g) on the moon is 1/6 th of that on the earth’s surface]
(a) \( \frac{1}{6} \text{ m} \)
(b) 6 m
(c) \( \frac{1}{36} \text{ m} \)
(d) \( \frac{1}{\sqrt{6}} \text{ m} \)
Answer: (a) \( \frac{1}{6} \text{ m} \)
In simple words: Since gravity on the moon is 6 times weaker than on Earth, a pendulum needs to be 6 times shorter to keep time the same way.

📝 Teacher's Note: Use the formula \( T = 2\pi\sqrt{L/g} \). Since the time period of a "seconds pendulum" is fixed at 2 seconds, length \( L \) must be directly proportional to gravity \( g \).

🎯 Exam Tip: "Seconds pendulum" always implies a time period of \( T = 2 \text{ seconds} \). Use the ratio method \( \frac{L_1}{L_2} = \frac{g_1}{g_2} \) to solve these quickly.

(iv) Two identical springs of constant k are connected, first in series and then in parallel. A metal block of mass m is suspended from their combination. The ratio of their frequencies of vertical oscillations will be in a ratio
(a) 1:4
(b) 1:2
(c) 2:1
(d) 4:1
Answer: (b) 1:2
In simple words: Connecting springs in a line (series) makes them "softer," while side-by-side (parallel) makes them "stiffer." Because frequency depends on stiffness, the side-by-side setup bounces twice as fast as the line setup.

📝 Teacher's Note: Explain that for series, \( k_s = k/2 \), and for parallel, \( k_p = 2k \). Since frequency \( f \propto \sqrt{k} \), the ratio becomes \( \sqrt{1/4} \).

🎯 Exam Tip: Remember: Series is like capacitors in parallel, and Parallel is like resistors in series. This mnemonic helps avoid mixing up spring constant formulas.

(v) The graph shows the variation of displacement of a particle performing S.H.M. with time t. Which of the following statements is correct from the graph?
(a) The acceleration is maximum at time T.
(b) The force is maximum at time 3T/4.
(c) The velocity is zero at time T/2.
(d) The kinetic energy is equal to total energy at time T/4.
Answer: (b) The force is maximum at time 3T/4.
In simple words: In a vibration, the force pulling the object back is strongest when the object is furthest from the middle. Looking at the graph, at time 3T/4, the object is at its maximum negative distance, so the force is at its peak.

📝 Teacher's Note: Use Newton's second law \( F = ma \) and the SHM relation \( a = -\omega^2 x \). Force magnitude is always maximum when displacement magnitude is maximum (at extreme positions).

🎯 Exam Tip: On a displacement-time graph, the peaks and troughs represent the maximum force and acceleration, while the x-intercepts (mean position) represent maximum velocity.

2. Answer in brief.

Question. Define linear simple harmonic motion.
Answer:
Definition: Linear simple harmonic motion (SHM) is defined as the linear periodic motion of a body, in which the force (or acceleration) is always directed towards the mean position and its magnitude is proportional to the displacement from the mean position.
OR
A particle is said to execute linear SHM if the particle undergoes oscillations about a point of stable equilibrium, subject to a linear restoring force always directed towards that point and whose magnitude is proportional to the magnitude of the displacement of the particle from that point.
Examples: The vibrations of the tines (prongs) of a tuning fork, the oscillations of the needle of a sewing machine.
In simple words: Simple harmonic motion is a back-and-forth movement where an object is always being pulled back to the center. The further away it goes, the harder it gets pulled back.

📝 Teacher's Note: Emphasize the "restoring" nature of the force. Students often forget to mention that the force is directed *towards* the mean position.

🎯 Exam Tip: Always include the mathematical condition \( F = -kx \) or \( a \propto -x \) when defining SHM to ensure you get full marks for technical accuracy.

Question. Using the differential equation of linear S.H.M, obtain the expression for (a) velocity in S.H.M., (b) acceleration in S.H.M.
Answer:
The general expression for the displacement of a particle in SHM at time t is \( x = A \sin (\omega t + \alpha) \) … (1)
where A is the amplitude, \( \omega \) is a constant in a particular case and \( \alpha \) is the initial phase.
The velocity of the particle is
\( v = \frac{dx}{dt} = \frac{d}{dt} [A \sin (\omega t + \alpha)] \)
\( \implies v = \omega A \cos (\omega t + \alpha) \)
\( \implies v = \omega A \sqrt{1 - \sin^2 (\omega t + \alpha)} \)
From Eq. (1), \( \sin (\omega t + \alpha) = x/A \)
\( \implies v = \omega A \sqrt{1 - \frac{x^2}{A^2}} \)
\( \implies v = \omega \sqrt{A^2 - x^2} \) … (2)
Equation (2) gives the velocity as a function of x.
The acceleration of the particle is
\( a = \frac{dv}{dt} = \frac{d}{dt} [A\omega \cos (\omega t + \alpha)] \)
\( \implies a = - \omega^2 A \sin (\omega t + \alpha) \)
But from Eq. (1), \( A \sin (\omega t + \alpha) = x \)
\( \implies a = -\omega^2 x \) … (3)
Equation (3) gives the acceleration as a function of x. The minus sign shows that the direction of the acceleration is opposite to that of the displacement.
In simple words: We can find how fast an object moves and how much it speeds up by looking at its position. Its speed depends on how far it is from the center, and its acceleration is always pushing it back toward the middle.

📝 Teacher's Note: This derivation starts from the displacement equation. Remind students that they can also start from the differential equation \( \frac{d^2x}{dt^2} + \omega^2x = 0 \) to derive these expressions by integration.

🎯 Exam Tip: Clearly state what each symbol represents (A, \( \omega \), \( \alpha \)) at the beginning of the derivation. Examiners look for the proper use of calculus notations.

Question. Obtain the expression for the period of a simple pendulum performing S.H.M.
Answer:
An ideal simple pendulum is defined as a heavy point mass suspended from a rigid support by a weightless, inextensible and twistless string, and set oscillating under gravity through a small angle in a vertical plane.
In practice, a small but heavy sphere, called the bob, is used. The distance from the point of suspension to the centre of gravity of the bob is called the length of the pendulum.
Consider a simple pendulum of length L suspended from a rigid support O. When displaced from its initial position of rest through a small angle \( \theta \) in a vertical plane and released, it performs oscillations between two extremes, B and C. At B, the forces on the bob are its weight \( m\vec{g} \) and the tension \( \vec{F_1} \) in the string. Resolve \( m\vec{g} \) into two components: \( mg \cos \theta \) in the direction opposite to that of the tension and \( mg \sin \theta \) perpendicular to the string.
\( mg \cos \theta \) is balanced by the tension in the string. \( mg \sin \theta \) restores the bob to the equilibrium position.
Restoring force, \( F = - mg \sin \theta \)
If \( \theta \) is small and expressed in radian,
\( \sin \theta \approx \theta = \frac{\text{arc}}{\text{radius}} = \frac{AB}{OB} = \frac{x}{L} \)
\( \implies F = - mg\theta = -mg \frac{x}{L} \) …. (1)
Since m, g and L are constant,
\( F \propto (-x) \) …. (2)
Thus, the net force on the bob is in the direction opposite to that of displacement x of the bob from its mean position, and its magnitude is proportional to the magnitude of the displacement. Hence, it follows that the motion of a simple pendulum is linear SHM.
Acceleration, \( a = \frac{F}{m} = -\frac{g}{L} x \) … (3)
Therefore, acceleration per unit displacement \( = |\frac{a}{x}| = \frac{g}{L} \) ….. (4)
Period of SHM,
\( T = \frac{2\pi}{\sqrt{\text{acceleration per unit displacement}}} \)
\( \implies T = \frac{2\pi}{\sqrt{g/L}} \)
\( \implies T = 2\pi\sqrt{\frac{L}{g}} \) … (5)
This gives the expression for the period of a simple pendulum.
In simple words: A pendulum swings back and forth because gravity pulls it down. By balancing the forces, we find that the time it takes for one full swing depends only on how long the string is and how strong gravity is.

📝 Teacher's Note: The "small angle approximation" (\( \sin\theta \approx \theta \)) is crucial. Explain to students that for large angles, the motion is still periodic but no longer "simple harmonic."

🎯 Exam Tip: Draw a neat, labeled diagram showing the resolution of weight. Clearly show the arc displacement \( x \) and its relation to the angle \( \theta \).

Question. State the laws of simple pendulum.
Answer:
The period of a simple pendulum at a given place is \( T = 2\pi\sqrt{\frac{L}{g}} \)
where L is the length of the simple pendulum and g is the acceleration due to gravity at that place. From the above expression, the laws of simple pendulum are as follows :
(1) Law of length : The period of a simple pendulum at a given place (g constant) is directly proportional to the square root of its length.
\( \implies T \propto \sqrt{L} \)
(2) Law of acceleration due to gravity : The period of a simple pendulum of a given length (L constant) is inversely proportional to the square root of the acceleration due to gravity.
\( \implies T \propto \frac{1}{\sqrt{g}} \)
(3) Law of mass : The period of a simple pendulum does not depend on the mass or material of the bob of the pendulum.
(4) Law of isochronism : The period of a simple pendulum does not depend on the amplitude of oscillations, provided that the amplitude is small.
In simple words: A pendulum's swing time changes with its length and gravity, but surprisingly, it doesn't care how heavy the weight is or how wide the swing is (as long as the swing isn't too huge).

📝 Teacher's Note: Many students intuitively think a heavier bob will swing faster. Use the "Law of Mass" to correct this misconception by showing mass cancels out in the acceleration equation.

🎯 Exam Tip: When stating these laws, always specify the conditions under which they hold true (e.g., "at a given place" or "for small amplitudes").

Question. Prove that under certain conditions a magnet vibrating in uniform magnetic field performs angular S.H.M.
Answer:
Consider a bar magnet of magnetic moment \( \mu \), suspended horizontally by a light twistless fibre in a region where the horizontal component of the Earth’s magnetic field is \( B_h \). The bar magnet is free to rotate in a horizontal plane. It comes to rest in approximately the North-South direction, along \( B_h \). If it is rotated in the horizontal plane by a small displacement \( \theta \) from its rest position (\( \theta = 0 \)), the suspension fibre is twisted. When the magnet is released, it oscillates about the rest position in angular or torsional oscillation.
The bar magnet experiences a torque \( \tau \) due to the field \( B_h \). Which tends to restore it to its original orientation parallel to \( B_h \). For small \( \theta \), this restoring torque is
\( \tau = – \mu B_h \sin \theta \approx – \mu B_h \theta \) …. (1)
where the minus sign indicates that the torque is opposite in direction to the angular displacement \( \theta \). Equation (1) shows that the torque (and hence the angular acceleration) is directly proportional in magnitude of the angular displacement but opposite in direction. Hence, for small angular displacement, the oscillations of the bar magnet in a uniform magnetic field is simple harmonic.
In simple words: When you twist a magnet in a magnetic field, the field pulls it back like a spring. Because this "magnetic pull" is proportional to the twist, the magnet vibrates in a perfect rhythmic pattern called SHM.

📝 Teacher's Note: Compare this to linear SHM (\( F = -kx \)). In angular SHM, torque replaces force and angular displacement replaces linear displacement (\( \tau = -c\theta \)).

🎯 Exam Tip: The keyword "small angular displacement" is essential here. Without it, the torque is proportional to \( \sin\theta \), not \( \theta \), and the motion wouldn't be "simple" harmonic.

Question 3. Obtain the expression for the period of a magnet vibrating in a uniform magnetic field and performing S.H.M.
Answer:
Definition: Angular SHM is defined as the oscillatory motion of a body in which the restoring torque responsible for angular acceleration is directly proportional to the angular displacement and its direction is opposite to that of angular displacement.
The differential equation of angular SHM is
\( I \frac{d^2\theta}{dt^2} + c\theta = 0 \) ….. (1)
where I = moment of inertia of the oscillating body, \( \frac{d^2\theta}{dt^2} \) = angular acceleration of the body when its angular displacement is \( \theta \), and c = torsion constant of the suspension wire,
\( \implies \frac{d^2\theta}{dt^2} + \frac{c}{I} \theta = 0 \)
Let \( \frac{c}{I} = \omega^2 \), a constant. Therefore, the angular frequency, \( \omega = \sqrt{c/I} \) and the angular acceleration,
\( a = \frac{d^2\theta}{dt^2} = -\omega^2\theta \) … (2)
The minus sign shows that the \( \alpha \) and \( \theta \) have opposite directions. The period T of angular SHM is
\( T = \frac{2\pi}{\omega} = \frac{2\pi}{\sqrt{c/I}} = 2\pi\sqrt{\frac{I}{c}} \) … (3)
This is the expression for the period in terms of torque constant. Also, from Eq. (2),
\( \omega = \sqrt{\frac{|\alpha|}{|\theta|}} = \sqrt{\text{angular acceleration per unit angular displacement}} \)
\( \implies T = \frac{2\pi}{\omega} = \frac{2\pi}{\sqrt{|\alpha/\theta|}} \)
\( \implies T = \frac{2\pi}{\sqrt{\text{angular acceleration per unit angular displacement}}} \)
In simple words: The time it takes for a magnet to swing back and forth depends on its weight distribution (Inertia) and how strongly the magnetic field pulls it. Heavier magnets or weaker fields make it swing slower.

📝 Teacher's Note: Connect this general formula back to the magnet by substituting the torque constant \( c = \mu B_h \). This shows how physical properties like magnetic moment directly affect the vibration time.

🎯 Exam Tip: In exams, sometimes they ask for the period in terms of magnetic moment. Be ready to substitute \( c = \mu B_h \) to get \( T = 2\pi\sqrt{\frac{I}{\mu B_h}} \).

Question 4. Show that a linear S.H.M. is the projection of a U.C.M. along any of its diameter.
Answer:
Consider a particle which moves anticlockwise around a circular path of radius A with a constant angular speed \( \omega \). Let the path lie in the x-y plane with the centre at the origin O. The instantaneous position P of the particle is called the reference point and the circle in which the particle moves as the reference circle.
The perpendicular projection of P onto the y-axis is Q. Then, as the particle travels around the circle, Q moves to-and-fro along the y-axis. Line OP makes an angle \( \alpha \) with the x-axis at t = 0. At time t, this angle becomes \( \theta = \omega t + \alpha \).
The projection Q of the reference point is described by the y-coordinate,
\( y = OQ = OP \sin \angle OPQ \), Since \( \angle OPQ = \omega t + \alpha \), \( y = A \sin(\omega t + \alpha) \)
which is the equation of a linear SHM of amplitude A. The angular frequency \( \omega \) of a linear SHM can thus be understood as the angular velocity of the reference particle.
The tangential velocity of the reference particle is \( v = \omega A \). Its y-component at time t is \( v_y = \omega A \sin (90^\circ – \theta) = \omega A \cos \theta \)
\( \implies v_y = \omega A \cos (\omega t + \alpha) \)
The centripetal acceleration of the reference particle is \( a_r = \omega^2 A \), so that its y-component at time t is \( a_x = a_r \sin \angle OPQ \)
\( \implies a_x = – \omega^2 A \sin (\omega t + \alpha) \)
In simple words: If you watch the shadow of a ball moving in a perfect circle, the shadow moves back and forth exactly like a spring would. This shows that circular motion and vibrating motion are deeply connected.

📝 Teacher's Note: This is an excellent way to visualize SHM. Use the analogy of a rotating wheel's shadow on a wall to help students "see" the projection moving linearly.

🎯 Exam Tip: Practice drawing the circle and the projection carefully. Label the y-coordinate as the SHM displacement and show the angle \( \omega t + \alpha \) clearly.

Question 5. Draw graphs of displacement, velocity and acceleration against phase angle, for a particle performing linear S.H.M. from (a) the mean position (b) the positive extreme position. Deduce your conclusions from the graph.
Answer:
Consider a particle performing SHM, with amplitude A and period \( T = 2\pi/\omega \).

Graphs Of SHM Starting From Mean Position

Starting from the mean position towards the positive extreme position where \( \omega \) is the angular frequency. Its displacement from the mean position \( (x) \), velocity \( (v) \) and acceleration \( (a) \) at any instant are
\( x = A \sin \omega t = A \sin \left( \frac{2\pi}{T} t \right) \) (\( \because \omega = \frac{2\pi}{T} \))
\( v = \frac{dx}{dt} = \omega A \cos \omega t = \omega A \cos \left( \frac{2\pi}{T} t \right) \)
\( a = \frac{dv}{dt} = -\omega^2 A \sin \omega t = -\omega^2 A \sin \left( \frac{2\pi}{T} t \right) \) as the initial phase \( \alpha = 0 \).
Using these expressions, the values of \( x \), \( v \) and \( a \) at the end of every quarter of a period, starting from \( t = 0 \), are tabulated below.

Using the values in the table we can plot graphs of displacement, velocity and acceleration with time.

Conclusions :

1. The displacement, velocity and acceleration of a particle performing linear SHM are periodic (harmonic) functions of time. For a particle starting at the mean position, the x-t and a-t graphs are sine curves. The v-t graph is a cosine curve.

2. There is a phase difference of \( \frac{\pi}{2} \) radians between x and v, and between v and a.

3. There is a phase difference of \( \pi \) radians between x and a.

Graphs Of SHM Starting From Positive Extreme Position

Consider a particle performing linear SHM with amplitude A and period \( T = 2\pi/\omega \), starting from the positive extreme position, where \( \omega \) is the angular frequency. Its displacement from the mean position \( (x) \), velocity \( (v) \) and acceleration \( (a) \) at any instant \( (t) \) are
\( x = A \cos \omega t = A \cos \left( \frac{2\pi}{T} t \right) \) (\( \because \omega = \frac{2\pi}{T} \))
\( v = -\omega A \sin \omega t = -\omega A \sin \left( \frac{2\pi}{T} t \right) \)
\( a = -\omega^2 A \cos \omega t = -\omega^2 A \cos \left( \frac{2\pi}{T} t \right) \)
Using these expressions, the values of \( x \), \( v \) and \( a \) at the end of every quarter of a period, starting from \( t = 0 \), are tabulated below.

Using these values, we can plot graphs showing the variation of displacement, velocity and acceleration with time.

Conclusions :

1. The displacement, velocity and acceleration of a particle performing linear SHM are periodic (harmonic) functions of time. For a particle starting from an extreme position, the x-t and a-t graphs are cosine curves; the v-t graph is a sine curve.

2. There is a phase difference of \( \frac{\pi}{2} \) radians between x and v, and between v and a.

3. There is a phase difference of \( \pi \) radians between x and a.

Explanations :
(1) v-t graph : It is a sine curve, i.e., the velocity is a periodic (harmonic) function of time which repeats after a phase of \( 2\pi \) rad. There is a phase difference of \( \pi/2 \) rad between a and v.
v is minimum (equal to zero) at the extreme positions (i.e., at \( x = \pm A \)) and v is maximum \( ( = \pm \omega A) \) at the mean position \( (x = 0) \).
(2) a-t graph : It is a cosine curve, i.e., the acceleration is a periodic (harmonic) function of time which repeats after a phase of \( 2\pi \) rad. There is a phase difference of \( \pi \) rad between v and a. a is minimum (equal to zero) at the mean position \( (x = 0) \) and a is maximum \( ( = \mp \omega^2 A) \) at the extreme positions \( (x = \pm A) \).

Question 6. Deduce the expressions for the kinetic energy and potential energy of a particle executing S.H.M. Hence obtain the expression for total energy of a particle performing S.H.M and show that the total energy is conserved. State the factors on which total energy depends.
Answer:
Consider a particle of mass m performing linear SHM with amplitude A. The restoring force acting on the particle is \( F = - kx \), where k is the force constant and x is the displacement of the particle from its mean position.
(1) Kinetic energy : At distance x from the mean position, the velocity is
\( v = \omega \sqrt{A^2 - x^2} \)
where \( \omega = \sqrt{k/m} \). The kinetic energy (KE) of the particle is
\( KE = \frac{1}{2} mv^2 = \frac{1}{2} m\omega^2 (A^2 - x^2) \)
\( = \frac{1}{2} k(A^2 - x^2) \dots (1) \)
If the phase of the particle at an instant t is \( \theta = \omega t + \alpha \), where \( \alpha \) is initial phase, its velocity at that instant is
\( v = \omega A \cos (\omega t + \alpha) \)
and its KE at that instant is
\( KE = \frac{1}{2} mv^2 = \frac{1}{2} m\omega^2 A^2 \cos^2 (\omega t + \alpha) \dots (2) \)
Therefore, the KE varies with time as \( \cos^2 \theta \).

(2) Potential energy : The potential energy of a particle in linear SHM is defined as the work done by an external agent, against the restoring force, in taking the particle from its mean position to a given point in the path, keeping the particle in equilibrium.
Suppose the particle is displaced from \( P_1 \) to \( P_2 \), through an infinitesimal distance dx against the restoring force F as shown.
The corresponding work done by the external agent will be \( dW = (- F)dx = kx \, dx \).
This work done is stored in the form of potential energy. The potential energy (PE) of the particle when its displacement from the mean position is x can be found by integrating the above expression from 0 to x.
\( \therefore PE = \int dW = \int_{0}^{x} kx \, dx = \frac{1}{2} kx^2 \dots (3) \)
The displacement of the particle at an instant t being
\( x = A \sin (\omega t + \alpha) \)
its PE at that instant is
\( PE = \frac{1}{2} kx^2 = \frac{1}{2} kA^2 \sin^2 (\omega t + \alpha) \dots (4) \)
Therefore, the PE varies with time as \( \sin^2 \theta \).

(3) Total energy : The total energy of the particle is equal to the sum of its potential energy and kinetic energy.
From Eqs. (1) and (3), total energy is \( E = PE + KE \)
\( = \frac{1}{2} kx^2 + \frac{1}{2} k(A^2 - x^2) \)
\( = \frac{1}{2} kx^2 + \frac{1}{2} kA^2 - \frac{1}{2} kx^2 \)

\( \implies E = \frac{1}{2} kA^2 = \frac{1}{2} m\omega^2 A^2 \dots (5) \)
As m is constant, \( \omega \) and A are constants of the motion, the total energy of the particle remains constant (or its conserved).
The total energy depends on mass (m), frequency (via \( \omega \)), and amplitude (A).
In simple words: In SHM, energy keeps changing between potential (energy of position) and kinetic (energy of motion), but their total sum always stays the same because the system is perfectly balanced.

📝 Teacher's Note: Use the analogy of a swing where at the highest point it has max potential energy and at the bottom max kinetic energy to explain conservation. Emphasize that while PE and KE change with position, their sum \( E \) is always \( \frac{1}{2} kA^2 \).

🎯 Exam Tip: When deriving total energy, show clearly how the \( x^2 \) terms cancel out. This proves that total energy is independent of the instantaneous displacement \( x \).

Question 7. Deduce the expression for period of simple pendulum. Hence state the factors on which its period depends.
Answer:
An ideal simple pendulum is defined as a heavy point mass suspended from a rigid support by a weightless, inextensible and twistless string, and set oscillating under gravity through a small angle in a vertical plane.
In practice, a small but heavy sphere, called the bob, is used. The distance from the point of suspension to the centre of gravity of the bob is called the length of the pendulum.
Consider a simple pendulum of length L suspended from a rigid support O. When displaced from its initial position of rest through a small angle \( \theta \) in a vertical plane and released, it performs oscillations between two extremes, B and C. At B, the forces on the bob are its weight \( mg \) and the tension \( F_1 \) in the string. Resolve \( mg \) into two components: \( mg \cos \theta \) in the direction opposite to that of the tension and \( mg \sin \theta \) perpendicular to the string.
\( mg \cos \theta \) balanced by the tension in the string. \( mg \sin \theta \) restores the bob to the equilibrium position.
Restoring force, \( F = - mg \sin \theta \)
If \( \theta \) is small and expressed in radian,
\( \sin \theta \approx \theta = \frac{\text{arc}}{\text{radius}} = \frac{AB}{OB} = \frac{x}{L} \)
\( \therefore F = - mg\theta = -mg \frac{x}{L} \dots (1) \)
Since m, g and L are constant,
\( F \propto (-x) \dots (2) \)
Thus, the net force on the bob is in the direction opposite to that of displacement x of the bob from its mean position, and the magnitude of the force is proportional to the magnitude of the displacement. Hence, it follows that the motion of a simple pendulum is linear SHM.
Acceleration, \( a = \frac{F}{m} = -\frac{g}{L} x \dots (3) \)
Therefore, acceleration per unit displacement
\( = \left| \frac{a}{x} \right| = \frac{g}{L} \dots (4) \)
Period of SHM,
\( T = \frac{2\pi}{\sqrt{\text{acceleration per unit displacement}}} \)

\( \implies T = \frac{2\pi}{\sqrt{g/L}} \)

\( \implies T = 2\pi \sqrt{\frac{L}{g}} \dots (5) \)
This gives the expression for the period of a simple pendulum. The laws of simple pendulum are:
(1) Law of length : \( T \propto \sqrt{L} \)
(2) Law of acceleration due to gravity : \( T \propto \frac{1}{\sqrt{g}} \)
In simple words: The time it takes for a pendulum to swing back and forth depends only on how long the string is and the gravity pulling on it; surprisingly, the weight of the bob doesn't change the timing at all.

📝 Teacher's Note: Use the small angle approximation (\( \sin \theta \approx \theta \)) to explain why the motion is SHM. If the angle is too large, the motion becomes complex and the simple period formula won't apply accurately.

🎯 Exam Tip: Always state the assumption "If \( \theta \) is small" before substituting \( \sin \theta \) with \( x/L \). Examiners look for this specific condition as it's the basis for linear SHM in a pendulum.

Question 8. At what distance from the mean position is the speed of a particle performing S.H.M. half its maximum speed. Given path length of S.H.M. = 10 cm. [Ans: 4.33 cm]
Answer:
Data: \( v = \frac{1}{2}v_{max} \), path length \( 2A = 10 \text{ cm} \)
\( \implies A = 5 \text{ cm} \)
\( v = \omega\sqrt{A^2 - x^2} \) and \( v_{max} = \omega A \)
Since \( v = \frac{1}{2}v_{max} \),
\( \omega\sqrt{A^2 - x^2} = \frac{\omega A}{2} \)

\( \implies A^2 - x^2 = \frac{A^2}{4} \)

\( \implies x^2 = A^2 - \frac{A^2}{4} = \frac{3A^2}{4} \)

\( \implies x = \pm \frac{\sqrt{3}}{2} A = \pm 0.866 \times 5 = \pm 4.33 \text{ cm} \)
This gives the required displacement.
In simple words: We are calculating the specific point in the object's path where it has slowed down to exactly half of its fastest speed. By comparing the speed formula to the maximum speed formula, we find this happens at roughly 4.33 cm from the center.

📝 Teacher's Note: Remind students that path length is twice the amplitude (\( 2A \)). A common mistake is to take the path length as the amplitude itself. Show them how the angular frequency \( \omega \) cancels out in the ratio, meaning this specific position depends only on the amplitude.

🎯 Exam Tip: When a question gives "path length," immediately divide it by 2 to get the amplitude. Clearly show the step where you square both sides to remove the square root sign to avoid calculation errors.

Question 9. In SI units, the differential equation of an S.H.M. is \( \frac{d^2x}{dt^2} = -36x \). Find its frequency and period. Find its frequency and period. [Ans: 0.9548 Hz, 1.047 s]
Answer:
\( \frac{d^2x}{dt^2} = -36x \)
Comparing this equation with the general equation,
\( \frac{d^2x}{dt^2} = -\omega^2 x \)
We get, \( \omega^2 = 36 \)

\( \implies \omega = 6 \text{ rad/s} \)
\( \omega = 2\pi f \)

\( \implies \) The frequency, \( f = \frac{\omega}{2\pi} = \frac{6}{2(3.142)} = \frac{6}{6.284} = 0.9548 \text{ Hz} \)
and the period, \( T = \frac{1}{f} = \frac{1}{0.9548} = 1.047 \text{ s} \)
In simple words: Every swinging object follows a math rule that links its acceleration to its position. By comparing the given rule to the standard one, we can figure out how many times it swings per second (frequency) and how long one full swing takes (period).

📝 Teacher's Note: Use this question to emphasize the pattern-matching technique in Physics. Students should memorize the standard differential equation of S.H.M. so they can easily identify \( \omega^2 \).

🎯 Exam Tip: Don't forget to write the units (Hz for frequency and s for period). Use the value of \( \pi \approx 3.142 \) for more accurate calculations as required in board exams.

Question 10. A needle of a sewing machine moves along a path of amplitude 4 cm with frequency 5 Hz. Find its acceleration \( \frac{1}{30} \) s after it has crossed the mean position. [Ans: 34.2 m/s\(^2\)]
Answer:
Data: \( A = 4 \text{ cm} = 4 \times 10^{-2} \text{ m} \), \( f = 5 \text{ Hz} \), \( t = \frac{1}{30} \text{ s} \)
\( \omega = 2\pi f = 2\pi(5) = 10\pi \text{ rad/s} \)
Therefore, the magnitude of the acceleration,
\( |a| = \omega^2 x = \omega^2 A \sin \omega t \)
\( = (10\pi)^2 (4 \times 10^{-2}) \sin(10\pi \cdot \frac{1}{30}) \)
\( = 100\pi^2 (0.04) \sin \frac{\pi}{3} \)
\( = 4 \pi^2 \sin \frac{\pi}{3} = 4 (9.872)(0.866) \approx 34.20 \text{ m/s}^2 \)
In simple words: The needle moves up and down very fast. We are calculating how hard it is "pushing" back toward the center exactly \( \frac{1}{30} \) of a second after it passes the middle.

📝 Teacher's Note: Explain that since the needle starts from the mean position, the displacement \( x \) is given by \( A \sin \omega t \). This helps students understand why we use the sine function for acceleration in this context.

🎯 Exam Tip: Ensure your calculator is in "Radian" mode when calculating \( \sin \frac{\pi}{3} \), or convert \( \frac{\pi}{3} \) to \( 60^\circ \) if using degree mode.

Question 11. Potential energy of a particle performing linear S.H.M is \( 0.1 \pi^2 x^2 \) joule. If mass of the particle is 20 g, find the frequency of S.H.M. [Ans: 1.581 Hz]
Answer:
Data: \( PE = 0.1 \pi^2 x^2 \text{ J} \), \( m = 20 \text{ g} = 2 \times 10^{-2} \text{ kg} \)
\( PE = \frac{1}{2} m \omega^2 x^2 = \frac{1}{2} m (4 \pi^2 f^2) x^2 \)

\( \implies \frac{1}{2} m (4 \pi^2 f^2) x^2 = 0.1 \pi^2 x^2 \)

\( \implies 2 m f^2 = 0.1 \)

\( \implies f^2 = \frac{0.1}{2(2 \times 10^{-2})} = \frac{0.1}{0.04} = 2.5 \)

\( \implies \) The frequency of SHM is \( f = \sqrt{2.5} = 1.581 \text{ Hz} \)
In simple words: Potential energy is "stored" energy. By knowing how much energy is stored at a certain distance and the weight of the object, we can calculate how fast it naturally vibrates.

📝 Teacher's Note: This is a classic "equating two formulas" problem. Help students see that since \( x^2 \) and \( \pi^2 \) appear on both sides, they cancel out, simplifying the math significantly.

🎯 Exam Tip: Always convert mass from grams to kilograms immediately. Missing the \( 10^{-3} \) factor is the most common cause of incorrect decimal places in the final answer.

Question 12. The total energy of a body of mass 2 kg performing S.H.M. is 40 J. Find its speed while crossing the centre of the path. [Ans: 6.324 m/s]
Answer:
Data: \( m = 2 \text{ kg} \), \( E = 40 \text{ J} \)
The speed of the body while crossing the centre of the path (mean position) is \( v_{max} \) and the total energy is entirely kinetic energy.

\( \implies \frac{1}{2} m v_{max}^2 = E \)

\( \implies v_{max} = \sqrt{\frac{2E}{m}} = \sqrt{\frac{2 \times 40}{2}} = \sqrt{40} = 6.324 \text{ m/s} \)
In simple words: At the very center of its swing, the object has no "stored" energy and is moving at its absolute fastest. All of its 40 Joules of energy is now energy of motion (kinetic energy).

📝 Teacher's Note: Use the Law of Conservation of Energy to explain why Total Energy = Max Kinetic Energy at the mean position. This makes the concept of \( v_{max} \) easier to grasp than just memorizing formulas.

🎯 Exam Tip: Note that the answer in the provided text says "cm/s" but the calculation results in "m/s". Always check your units based on the given SI values (\( kg \) and \( J \)).

Question 13. A simple pendulum performs S.H.M of period 4 seconds. How much time after crossing the mean position, will the displacement of the bob be one third of its amplitude. [Ans: 0.2163 s]
Answer:
Data: \( T = 4 \text{ s} \), \( x = \frac{A}{3} \)
The displacement of a particle starting into SHM from the mean position is \( x = A \sin \omega t = A \sin \frac{2\pi}{T} t \)

\( \implies A \sin \frac{2\pi}{T} t = \frac{A}{3} \)

\( \implies \frac{2\pi}{T} t = \sin^{-1} 0.3333 = 19.47^\circ = 19.47 \times \frac{\pi}{180} \text{ rad} \)

\( \implies \frac{2}{4} t = \frac{19.47}{180} \)

\( \implies t = \frac{19.47}{90} = 0.2163 \text{ s} \)

\( \implies \) the displacement of the bob will be one-third of its amplitude 0.2163 s after crossing the mean position.
In simple words: We are timing the pendulum. We want to know how long it takes to travel from the center to a point that is \( 1/3 \) of the way to the end of its swing.

📝 Teacher's Note: Students often struggle with \( \sin^{-1} \) values. Encourage them to use scientific calculators or log tables correctly. Explain the conversion from degrees to radians as it's crucial for the formula.

🎯 Exam Tip: When the object starts from the "mean position," always use the \( \sin \) function. If it starts from the "extreme position," use the \( \cos \) function.

Question 14. A simple pendulum of length 100 cm performs S.H.M. Find the restoring force acting on its bob of mass 50 g when the displacement from the mean position is 3 cm. [Ans: 1.48 \( \times 10^{-2} \) N]
Answer:
Data: \( L = 100 \text{ cm} = 1 \text{ m} \), \( m = 50 \text{ g} = 5 \times 10^{-2} \text{ kg} \), \( x = 3 \text{ cm} = 0.03 \text{ m} \), \( g = 9.8 \text{ m/s}^2 \)
Restoring force, \( F = mg \sin \theta \approx mg\theta = mg \frac{x}{L} \)
\( = (5 \times 10^{-2})(9.8)(\frac{3}{100}) \)
\( = 1.47 \times 10^{-2} \text{ N} \)
In simple words: Gravity pulls the pendulum bob back toward the center. This "pull" is what we call the restoring force, and it gets stronger as the bob moves further away from the middle.

📝 Teacher's Note: Explain the "small angle approximation" where \( \sin \theta \approx \theta \approx \frac{x}{L} \). This is why the formula for a simple pendulum works linearly for small displacements.

🎯 Exam Tip: Keep all measurements in meters and kilograms. Here, converting \( 100 \text{ cm} \) to \( 1 \text{ m} \) and \( 3 \text{ cm} \) to \( 0.03 \text{ m} \) prevents common power-of-ten errors.

Question 15. Find the change in length of a second’s pendulum, if the acceleration due to gravity at the place changes from 9.75 m/s\(^2\) to 9.80 m/s\(^2\). [Ans: Decreases by 5.07 mm]
Answer:
Data: \( g_1 = 9.75 \text{ m/s}^2 \), \( g_2 = 9.8 \text{ m/s}^2 \)
Length of a seconds pendulum, \( L = \frac{g}{\pi^2} \)

\( \implies L_1 = \frac{g_1}{\pi^2} = \frac{9.75}{9.872} = 0.9876 \text{ m} \)
and \( L_2 = \frac{g_2}{\pi^2} = \frac{9.8}{9.872} = 0.9927 \text{ m} \)

\( \implies \) The length of the seconds pendulum must be increased from 0.9876 m to 0.9927 m, i.e., by 0.0051 m (or 5.1 mm).
In simple words: A "seconds pendulum" is one that takes exactly 2 seconds for a full swing. If gravity gets slightly stronger, the string must be made slightly longer to keep the swing time the same.

📝 Teacher's Note: Define a seconds pendulum clearly: it's a pendulum with a period \( T = 2 \) seconds. Use the formula \( T = 2\pi\sqrt{L/g} \) to derive \( L = g/\pi^2 \).

🎯 Exam Tip: Read the question carefully. The provided answer key in the image says "Decreases", but the calculation shows an increase. Trust your calculated steps: as \( g \) increases, \( L \) must also increase to keep the ratio constant.

Question 16. At what distance from the mean position is the kinetic energy of a particle performing S.H.M. of amplitude 8 cm, three times its potential energy? [Ans: 4 cm]
Answer:
Data: \( A = 8 \text{ cm} \), \( KE = 3 PE \)
\( KE = \frac{1}{2} k(A^2 - x^2) \) and \( PE = \frac{1}{2} kx^2 \)
Given, \( KE = 3PE \).

\( \implies \frac{1}{2} k(A^2 - x^2) = 3(\frac{1}{2} kx^2) \)

\( \implies A^2 - x^2 = 3x^2 \)

\( \implies 4x^2 = A^2 \)

\( \implies \) the required displacement is \( x = \pm \frac{A}{2} = \pm \frac{8}{2} = \pm 4 \text{ cm} \)
In simple words: We are looking for the spot where the energy of motion is three times larger than the energy stored. It turns out this happens exactly halfway between the center and the furthest point.

📝 Teacher's Note: This question helps students visualize energy distribution. At the mean (\( x=0 \)), all is KE. At extreme (\( x=A \)), all is PE. Halfway point (\( x=A/2 \)) is a special point where KE is much larger than PE.

🎯 Exam Tip: Problems involving ratios of energies are common. Always write out the full expressions for KE and PE first so you can cancel out the constants like \( \frac{1}{2} \) and \( k \).

Question 17. A particle performing linear S.H.M. of period \( 2\pi \) seconds about the mean position O is observed to have a speed of \( b\sqrt{3} \) m/s, when at a distance \( b \) (metre) from O. If the particle is moving away from O at that instant, find the time required by the particle, to travel a further distance \( b \). [Ans: \( \pi/3 \) s]
Answer:
Data: \( T = 2\pi \text{ s} \), \( v = b\sqrt{3} \text{ m/s} \) at \( x = b \)

\( \implies \omega = \frac{2\pi}{T} = \frac{2\pi}{2\pi} = 1 \text{ rad/s} \)
\( v = \omega \sqrt{A^2 - x^2} \)

\( \implies \) At \( x=b \), \( b\sqrt{3} = (1)\sqrt{A^2 - b^2} \)

\( \implies 3b^2 = A^2 - b^2 \)

\( \implies A = 2b \)
Assuming the particle starts from the mean position, its displacement is given by
\( x = A \sin \omega t = 2b \sin t \)
If the particle is at \( x = b \) at \( t = t_1 \),
\( b = 2b \sin t_1 \)

\( \implies t_1 = \sin^{-1} \frac{1}{2} = \frac{\pi}{6} \text{ s} \)
Also, with period \( T = 2\pi \text{ s} \), on travelling a further distance \( b \) (total \( x = b+b = 2b = A \)) the particle will reach the positive extremity at time \( t_2 = \frac{T}{4} = \frac{2\pi}{4} = \frac{\pi}{2} \text{ s} \).

\( \implies \) The time taken to travel a further distance \( b \) from \( x = b \) is \( t_2 - t_1 = \frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3} \text{ s} \).
In simple words: We first use the speed to find out how far the object can swing (its amplitude). Then we calculate how long it takes to go from one point to another point further along its path.

📝 Teacher's Note: This is a multi-step problem. First find \( \omega \), then \( A \), then use the displacement equation. Point out that since \( A=2b \), a "further distance \( b \)" means reaching the extreme position.

🎯 Exam Tip: Be careful with the phrase "further distance". It means you need to calculate the time for the new total displacement and subtract the time for the initial displacement.

Question 18. The period of oscillation of a body of mass \( m_1 \) suspended from a light spring is \( T \). When a body of mass \( m_2 \) is tied to the first body and the system is made to oscillate, the period is \( 2T \). Compare the masses \( m_1 \) and \( m_2 \) [Ans: 1/3]
Answer:
\( T = 2\pi\sqrt{\frac{m_1}{k}} \)

\( \implies \frac{2T}{T} = 2 = \sqrt{\frac{m_1 + m_2}{m_1}} \)

\( \implies \frac{m_1 + m_2}{m_1} = 4 \)

\( \implies \frac{m_2}{m_1} = 3 \)

\( \implies \frac{m_1}{m_2} = \frac{1}{3} \)
This gives the required ratio of the masses.
In simple words: Adding mass to a spring makes it swing slower. If adding a second mass doubles the time it takes to swing, the second mass must be three times heavier than the first one.

📝 Teacher's Note: Use the proportionality \( T \propto \sqrt{m} \). Show that doubling the period means the total mass must have quadrupled (\( 2^2 = 4 \)).

🎯 Exam Tip: "Compare the masses" usually means finding the ratio \( m_1 : m_2 \). Always double-check if the question asks for \( m_1/m_2 \) or \( m_2/m_1 \).

Question 19. The displacement of an oscillating particle is given by \( x = a \sin \omega t + b \cos \omega t \) where \( a, b \) and \( \omega \) are constants. Prove that the particle performs a linear S.H.M. with amplitude \( A = \sqrt{a^2 + b^2} \)
Answer:
\( x = a \sin \omega t + b \cos \omega t \)
Let \( a = A \cos \phi \) and \( b = A \sin \phi \), so that
\( A^2 = a^2 + b^2 \) and \( \tan \phi = \frac{b}{a} \)

\( \implies x = A \cos \phi \sin \omega t + A \sin \phi \cos \omega t \)

\( \implies x = A \sin(\omega t + \phi) \)
which is the equation of a linear SHM with amplitude \( A = \sqrt{a^2 + b^2} \) and phase constant \( \phi = \tan^{-1} \frac{b}{a} \), as required.
In simple words: Combining two different types of waves (sine and cosine) of the same speed creates a new single wave. We prove that this combined wave is just a regular swing with a specific total height.

📝 Teacher's Note: This is a proof of the superposition of two perpendicular SHMs or the addition of two functions. It's a great application of trigonometric identity \( \sin(A+B) \).

🎯 Exam Tip: In proofs, clearly state your substitutions (like \( a = A \cos \phi \)) as these are the keys to simplifying the expression into the standard SHM form.

Question 20. Two parallel S.H.M.s represented by \( x_1 = 5 \sin(4\pi t + \frac{\pi}{3}) \text{ cm} \) and \( x_2 = 3 \sin(4\pi t + \pi/4) \text{ cm} \) are superposed on a particle. Determine the amplitude and epoch of the resultant S.H.M. [Ans: 7.936 cm, 54\(^\circ\) 23']
Answer:
Data: \( x_1 = 5 \sin(4\pi t + \frac{\pi}{3}) = A_1 \sin(\omega t + \alpha) \),
\( x_2 = 3 \sin(4\pi t + \frac{\pi}{4}) = A_2 \sin(\omega t + \beta) \)

\( \implies A_1 = 5 \text{ cm}, A_2 = 3 \text{ cm}, \alpha = \frac{\pi}{3} \text{ rad}, \beta = \frac{\pi}{4} \text{ rad} \)
(i) Resultant amplitude,
\( R = \sqrt{A_1^2 + A_2^2 + 2A_1 A_2 \cos(\alpha - \beta)} \)
\( = \sqrt{(5)^2 + (3)^2 + 2(5)(3) \cos(\frac{\pi}{3} - \frac{\pi}{4})} \)
\( = \sqrt{25 + 9 + 30 \cos \frac{\pi}{12}} = \sqrt{34 + 30(0.9659)} \)
\( = \sqrt{34 + 28.98} = \sqrt{62.98} = 7.936 \text{ cm} \)
(ii) Epoch of the resultant SHM,
\( \delta = \tan^{-1} \frac{A_1 \sin \alpha + A_2 \sin \beta}{A_1 \cos \alpha + A_2 \cos \beta} \)
\( = \tan^{-1} \frac{5 \sin(\pi/3) + 3 \sin(\pi/4)}{5 \cos(\pi/3) + 3 \cos(\pi/4)} \)
\( = \tan^{-1} \frac{5(0.866) + 3(0.7071)}{5(0.5) + 3(0.7071)} \)
\( = \tan^{-1} \frac{4.33 + 2.1213}{2.5 + 2.1213} = \tan^{-1} 1.396 = 54^\circ 23' \)
In simple words: When two similar motions push an object at the same time, they combine. We are finding how big the final total swing is (amplitude) and exactly where it "starts" its cycle (epoch).

📝 Teacher's Note: Compare this to vector addition. The resultant amplitude formula is identical to the law of cosines for vectors. This makes it easier for students who are already comfortable with vectors.

🎯 Exam Tip: The term "epoch" just refers to the initial phase angle \( \delta \). Use accurate values for \( \sin 60^\circ \) and \( \sin 45^\circ \) to get the correct result in degrees and minutes.

Question 21. A 20 cm wide thin circular disc of mass 200 g is suspended to a rigid support from a thin metallic string. By holding the rim of the disc, the string is twisted through 60\(^\circ\) and released. It now performs angular oscillations of period 1 second. Calculate the maximum restoring torque generated in the string under undamped conditions. (\( \pi^3 \approx 31 \)) [Ans: 0.04133 N m]
Answer:
Data: \( R = 10 \text{ cm} = 0.1 \text{ m} \), \( M = 0.2 \text{ kg} \), \( \theta_m = 60^\circ = \frac{\pi}{3} \text{ rad} \), \( T = 1 \text{ s} \), \( \pi^3 \approx 31 \)
The MI of the disc about the rotation axis (perpendicular through its centre) is
\( I = \frac{1}{2}MR^2 = (0.1)(0.1)^2 = 10^{-3} \text{ kg.m}^2 \)
The period of torsional oscillation, \( T = 2\pi\sqrt{\frac{I}{c}} \)

\( \implies \) The torsion constant, \( c = 4\pi^2 \frac{I}{T^2} \)
The magnitude of the maximum restoring torque,
\( \tau_{max} = c\theta_m = (4\pi^2 \frac{I}{T^2})(\frac{\pi}{3}) \)
\( = \frac{4}{3}\pi^3 \frac{I}{T^2} = \frac{4}{3} (31) (\frac{10^{-3}}{1^2}) \)
\( = 41.33 \times 10^{-3} = 0.04133 \text{ N.m} \)
In simple words: When you twist a hanging disc and let go, the string tries to untwist. We are calculating the maximum "turning force" (torque) the string uses to pull the disc back.

📝 Teacher's Note: Differentiate between linear SHM (\( F = -kx \)) and angular SHM (\( \tau = -c\theta \)). Emphasize that \( I \) here is for a disc rotating around its central axis.

🎯 Exam Tip: Note the given approximation \( \pi^3 \approx 31 \). Using this specific value will make the final arithmetic easier and lead exactly to the expected answer.

Question 22. Find the number of oscillations performed per minute by a magnet is vibrating in the plane of a uniform field of \( 1.6 \times 10^{-5} \text{ Wb/m}^2 \). The magnet has moment of inertia \( 3 \times 10^{-6} \text{ kgm}^2 \) and magnetic moment \( 3 \text{ A m}^2 \). [Ans: 38.19 osc/min.]
Answer:
Data: \( B = 1.6 \times 10^{-5} \text{ T} \), \( I = 3 \times 10^{-6} \text{ kg.m}^2 \), \( \mu = 3 \text{ A.m}^2 \)
The period of oscillation, \( T = 2\pi\sqrt{\frac{I}{\mu B}} \)

\( \implies \) The frequency of oscillation is \( f = \frac{1}{2\pi}\sqrt{\frac{\mu B}{I}} \)

\( \implies \) The number of oscillations per minute \( = 60f \)
\( = 60 \cdot \frac{1}{2\pi} \sqrt{\frac{3(1.6 \times 10^{-5})}{3 \times 10^{-6}}} = \frac{60}{2\pi} \sqrt{16} = \frac{120}{3.142} \)
\( = 38.19 \text{ per minute} \)
In simple words: A magnet in a magnetic field acts like a compass needle. If you nudge it, it vibrates. We are counting how many times it wiggles back and forth in one minute.

📝 Teacher's Note: This is an application of angular SHM in magnetism. Remind students that the formula is very similar to the pendulum formula, replacing mechanical terms with magnetic ones (\( \mu \) and \( B \)).

🎯 Exam Tip: The question asks for oscillations per minute, but the formula gives frequency in Hz (per second). Don't forget to multiply your frequency by 60 at the end.

Question 23. A wooden block of mass m is kept on a piston that can perform vertical vibrations of adjustable frequency and amplitude. During vibrations, we don't want the block to leave the contact with the piston. How much maximum frequency is possible if the amplitude of vibrations is restricted to 25 cm? In this case, how much is the energy per unit mass of the block? (\( g \approx \pi^2 \approx 10 \text{ m s}^{-2} \)) [Ans: \( n_{max} = 1/s \), \( E/m = 1.25 \text{ J/kg} \)]
Answer:
Data: \( A = 0.25 \text{ m} \), \( g = \pi^2 = 10 \text{ m/s}^2 \)
During vertical oscillations, the acceleration is maximum at the turning points at the top and bottom. The block will just lose contact with the piston when its apparent weight is zero at the top, i. e., when its acceleration is \( a_{max} = g \), downwards.
\( |a_{max}| = \omega^2 A = 4\pi^2 f^2 A \)

\( \implies 4\pi^2 f_{max}^2 A = g \)

\( \implies f_{max} = \sqrt{\frac{g}{4\pi^2 A}} = \sqrt{\frac{10}{4(10)(0.25)}} = 1 \text{ Hz} \)
This gives the required frequency of the piston.
\( E = \frac{1}{2} m \omega^2 A^2 = \frac{1}{2} m (4\pi^2 f^2) A^2 \)

\( \implies \frac{E}{m} = 2\pi^2 f^2 A^2 = 2(10)(1)^2 (\frac{1}{4})^2 \)
\( = \frac{20}{16} = \frac{5}{4} = 1.25 \text{ J/kg} \)
In simple words: If the piston moves down too fast, the block will "float" off it. We are finding the limit where the piston accelerates at the same rate as gravity. We also calculate the energy contained in every kilogram of that block.

📝 Teacher's Note: This is a great real-world application. Explain that at the top, if the downward acceleration exceeds \( g \), the piston drops faster than the block can fall, causing separation.

🎯 Exam Tip: The condition for losing contact is always \( \omega^2 A = g \). Memorizing this condition for vertical SHM problems saves time during exams.

Can you tell? (Textbook Page No. 112)

Question. Why is the term angular frequency (\( \omega \)) used here for a linear motion?
Answer:
A linear SHM is the projection of a UCM (Uniform Circular Motion) on the diameter of the circle. The angular speed \( \omega \) of a particle moving along this reference circle is called the angular frequency of the particle executing linear SHM.
In simple words: Even though the object moves in a straight line, we can imagine it as a shadow of something spinning in a circle. The "spinning speed" of that imaginary circle is what we call angular frequency.

📝 Teacher's Note: Use a flashlight and a spinning ball to demonstrate how a circular motion creates a linear shadow. This visual aid makes the concept of "reference circle" concrete.

🎯 Exam Tip: Remember the phrase "projection of UCM on a diameter". Examiners look for this specific technical description.

Can you tell? (Textbook Page No. 114)

Question 1. State at which point during an oscillation the oscillator has zero velocity but positive acceleration?
Answer:
At the left extreme, i.e., \( x = -A \), so that \( a = -\omega^2 x = -\omega^2(-A) = \omega^2 A = a_{max} \).
In simple words: At the very far left end of a swing, the object stops for a tiny split second (zero velocity) and is about to be pulled back to the right as hard as possible (positive acceleration).

📝 Teacher's Note: Contrast this with the mean position, where velocity is maximum but acceleration is zero. It helps to draw a graph of \( v \) vs \( x \) and \( a \) vs \( x \).

🎯 Exam Tip: Be careful with signs. Negative displacement (\( -A \)) results in positive acceleration because the force is always directed toward the center (\( x=0 \)).

Question 2. During which part of the simple harmonic motion velocity is positive but the displacement is negative, and vice versa?
Answer:
Velocity \( v \) is positive (to the right) while displacement \( x \) is negative when the particle in SHM is moving from the left extreme towards the mean position.
Velocity \( v \) is negative (to the left) while displacement \( x \) is positive when the particle in SHM is moving from the right extreme towards the mean position.
In simple words: This happens when the object is on one side of the center but is already heading back toward the middle.

📝 Teacher's Note: Use a simple diagram with arrows to show the direction of motion relative to the center. This helps students visualize phase differences.

🎯 Exam Tip: Remember: when moving "towards the center," the signs of velocity and displacement are always opposite.

Can you tell? (Textbook page 76)

Question 1. To start a pendulum swinging, usually you pull it slightly to one side and release it. What kind of energy is transferred to the mass in doing this?
Answer:
On pulling the bob of a simple pendulum slightly to one side, it is raised to a slightly higher position. Thus, it gains gravitational potential energy.
In simple words: By pulling it aside, you are actually lifting it up a little bit. This gives it "height energy" that will make it swing when you let go.

📝 Teacher's Note: This is a simple conservation of energy introduction. Remind students that work done by the hand is stored as PE.

🎯 Exam Tip: Always mention "gravitational" before potential energy for pendulums to show you understand the source of the energy.

Question 2. Describe the energy changes that occur when the mass is released.
Answer:
When released, the bob oscillates in SHM in a vertical plane and the energy oscillates back and forth between kinetic and potential, going completely from one form of energy to the other as the pendulum oscillates. In the case of undamped SHM, the motion starts with all of the energy as gravitational potential energy. As the object starts to move, the gravitational potential energy is converted into kinetic energy, becoming entirely kinetic energy at the equilibrium position. The velocity becomes zero at the other extreme as the kinetic energy is completely converted back into gravitational potential energy, and this cycle then repeats.
In simple words: The energy swaps places constantly. It's all height at the ends, and all speed in the middle.

📝 Teacher's Note: Use a "Total Energy = PE + KE" bar chart analogy to show how one grows as the other shrinks, while the total stays constant.

🎯 Exam Tip: Key phrases to include are "equilibrium position" (where KE is max) and "extreme positions" (where PE is max).

Question 3. Is/are there any other way/ways to start the oscillations of a pendulum? Which energy is supplied in this case/cases?
Answer:
The bob can be given a kinetic energy at its equilibrium position or at any other position of its path. In the first case, the motion starts with all of the energy as kinetic energy. In the second case, the motion starts with partly gravitational potential energy and partly kinetic energy.
In simple words: Instead of pulling and dropping it, you could just flick it while it's hanging still. That gives it speed (kinetic energy) right away to start its journey.

📝 Teacher's Note: This illustrates that the starting conditions (initial displacement and initial velocity) determine the total energy of the SHM system.

🎯 Exam Tip: Mentioning that energy can be supplied as "initial velocity at the mean position" is a technically accurate way to describe "flicking" the bob.

Can you tell? (Textbook Page No. 109)

Question 1. Is the motion of a leaf of a tree blowing in the wind periodic ?
Answer:
The leaf of a tree blowing in the wind oscillates, but the motion is not periodic. Also, its displacement from the equilibrium position is not a regular function of time.
In simple words: No. Even though the leaf wiggles back and forth, it doesn't do it in a perfectly timed, predictable rhythm because the wind changes constantly.

📝 Teacher's Note: Distinguish between "oscillatory" (back and forth) and "periodic" (happening at fixed intervals). All SHM is periodic, but not all oscillations are SHM.

🎯 Exam Tip: To be periodic, a motion must repeat itself exactly in equal intervals of time. Use this definition to justify why irregular motions like a leaf in the wind are not periodic.