JEE Physics Oscillations MCQs Set 08

Answer: (a) \( \dfrac{\sqrt{2.5}}{\pi}\ \text{s}^{-1} \)

Explanation:
1. In SHM \( |a|=\omega^{2}x \), so \( \omega^{2}=\dfrac{20}{2}=10\ \text{s}^{-2} \) and \( \omega=\sqrt{10} \) rad/s.
2. \( f=\dfrac{\omega}{2\pi}=\dfrac{\sqrt{10}}{2\pi}=\dfrac{\sqrt{2.5}}{\pi} \) (since \( \sqrt{10}=2\sqrt{2.5} \)).

Teacher's Note:
1. The slope of the a-x graph has magnitude \( \omega^{2} \).
2. Divide the angular frequency by 2π to get the frequency.

Answer: (b) y = 0.8 cos 10πt

Explanation:
1. \( \omega=\dfrac{2\pi}{T}=\dfrac{2\pi}{0.2}=10\pi \) rad/s.
2. \( y=A\sin(\omega t+\phi)=0.8\sin\left(10\pi t+\dfrac{\pi}{2}\right) \).
3. Since \( \sin(\theta+\pi/2)=\cos\theta \), \( y=0.8\cos10\pi t \).

Teacher's Note:
1. First find ω from the time period, then insert the initial phase.
2. A phase of π/2 turns a sine into a cosine.

Answer: (c) \( K_2=2K_1 \)

Explanation:
1. Maximum speed \( v_{max}=\omega A=\sqrt{\dfrac{g}{l}}\,(l\theta_0)=\theta_0\sqrt{gl} \).
2. \( K_{max}=\dfrac{1}{2}mv_{max}^{2}=\dfrac{1}{2}mgl\,\theta_0^{2} \), which is proportional to l for the same m and \( \theta_0 \).
3. Doubling l doubles the kinetic energy: \( K_2=2K_1 \).

Teacher's Note:
1. For a pendulum the amplitude is \( l\theta_0 \), so the length enters both ω and A.
2. The maximum kinetic energy equals the maximum potential energy \( mgl(1-\cos\theta_0) \).

Answer: (c) 3 : 1

Explanation:
1. Starting from equilibrium, \( x=A\sin\dfrac{2\pi t}{T} \). At \( t=\dfrac{T}{12} \) the phase is \( \dfrac{2\pi}{12}=30^\circ \).
2. \( \text{KE}\propto\cos^{2}30^\circ=\dfrac{3}{4} \) and \( \text{PE}\propto\sin^{2}30^\circ=\dfrac{1}{4} \).
3. KE : PE = 3 : 1.

Teacher's Note:
1. Kinetic energy is \( \dfrac{1}{2}m\omega^{2}A^{2}\cos^{2}\omega t \) and potential energy is \( \dfrac{1}{2}m\omega^{2}A^{2}\sin^{2}\omega t \) when the motion starts at equilibrium.
2. Total energy stays constant during the motion.

Answer: (b) \( \left(\dfrac{2k_0}{m\omega^{2}}\right)^{1/2}\sin\omega t \)

Explanation:
1. The kinetic energy is maximum at t = 0, so the particle is at the mean position then: \( x=A\sin\omega t \).
2. Maximum kinetic energy \( k_0=\dfrac{1}{2}m\omega^{2}A^{2} \), so \( A=\sqrt{\dfrac{2k_0}{m\omega^{2}}} \).
3. \( x=\left(\dfrac{2k_0}{m\omega^{2}}\right)^{1/2}\sin\omega t \).

Teacher's Note:
1. Maximum kinetic energy equals the total energy \( \dfrac{1}{2}m\omega^{2}A^{2} \).
2. Check dimensions: \( k_0/(m\omega^{2}) \) has the dimension of length squared.

Answer: (c) \( \dfrac{2\pi}{3}\sqrt{\dfrac{m}{k}} \)

Explanation:
1. The body oscillates about the natural-length position with amplitude \( A=2x_0 \), starting from the extreme compressed position \( x=-A \).
2. It strikes the wall when \( x=+x_0=+\dfrac{A}{2} \): \( -A\cos\omega t=\dfrac{A}{2} \), so \( \cos\omega t=-\dfrac{1}{2} \) and \( \omega t=\dfrac{2\pi}{3} \).
3. \( t=\dfrac{2\pi}{3}\sqrt{\dfrac{m}{k}} \) (using \( \omega=\sqrt{k/m} \)).

Teacher's Note:
1. Take the extreme position as the starting point and write \( x=-A\cos\omega t \).
2. The time from an extreme to the centre is T/4 and from the centre to half the amplitude is T/12, giving T/3 in total.

Answer: (b) \( 2\pi\sqrt{\dfrac{L}{g\cos\alpha}} \)

Explanation:
1. The vehicle accelerates down the incline with \( a=g\sin\alpha \). In the vehicle's frame the pendulum feels gravity plus a pseudo force \( -ma \) up the incline.
2. The component of g along the incline is cancelled by the pseudo force, leaving \( g_{eff}=g\cos\alpha \) perpendicular to the incline.
3. \( T=2\pi\sqrt{\dfrac{L}{g\cos\alpha}} \).

Teacher's Note:
1. The effective acceleration is the vector sum of g and the negative of the frame's acceleration.
2. When \( \alpha=0 \) the formula reduces to the usual \( 2\pi\sqrt{L/g} \).

Answer: (d) \( 2\sqrt2 \) cm

Explanation:
1. Kinetic energy equals half the total energy, so potential energy is also half: \( \dfrac{1}{2}k x^{2}=\dfrac{1}{2}\left(\dfrac{1}{2}kA^{2}\right) \).
2. \( x^{2}=\dfrac{A^{2}}{2} \), so \( x=\dfrac{A}{\sqrt2}=\dfrac{4}{\sqrt2} \).
3. \( x=2\sqrt2 \) cm.

Teacher's Note:
1. At \( x=\pm A/\sqrt2 \) the kinetic and potential energies are equal.
2. The displacement is always less than the amplitude, which rules out the 10 cm option.

Answer: (c) \( \dfrac{16}{9} \)

Explanation:
1. \( T=2\pi\sqrt{\dfrac{M}{k}} \) and \( \dfrac{5T}{3}=2\pi\sqrt{\dfrac{M+m}{k}} \).
2. Dividing: \( \dfrac{5}{3}=\sqrt{\dfrac{M+m}{M}} \), so \( \dfrac{M+m}{M}=\dfrac{25}{9} \).
3. \( \dfrac{m}{M}=\dfrac{25}{9}-1=\dfrac{16}{9} \).

Teacher's Note:
1. The time period is proportional to the square root of the mass.
2. Square the ratio of the periods before comparing the masses.

Answer: (c) \( -\dfrac{\pi}{6} \)

Explanation:
1. \( y_2=0.1\cos100\pi t=0.1\sin\left(100\pi t+\dfrac{\pi}{2}\right) \).
2. The velocity of an SHM leads its displacement by the same phase \( \dfrac{\pi}{2} \) for both particles, so the phase difference of velocities equals that of displacements.
3. \( \phi_1-\phi_2=\dfrac{\pi}{3}-\dfrac{\pi}{2}=-\dfrac{\pi}{6} \).

Teacher's Note:
1. Express both motions as sines (or both as cosines) before comparing phases.
2. Velocity and displacement are always \( \pi/2 \) apart in phase, so differentiating does not change the phase difference.

Answer: (c) \( \dfrac{\pi}{10} \) s

Explanation:
1. The object leaves the platform if the downward acceleration of the platform exceeds g. The maximum acceleration is \( \omega^{2}A \).
2. Condition: \( \omega^{2}A\le g \), so \( \omega\le\sqrt{\dfrac{10}{0.025}}=20 \) rad/s.
3. Least period \( T=\dfrac{2\pi}{20}=\dfrac{\pi}{10} \) s.

Teacher's Note:
1. The object stays in contact as long as the maximum acceleration of the platform does not exceed g.
2. A larger frequency means a smaller period, so the limiting frequency gives the least period.

Answer: (d) \( x_0\omega \)

Explanation:
1. \( x=\dfrac{x_0}{2}(1+\cos2\omega t) \), so the mean position is \( x=\dfrac{x_0}{2} \) and the amplitude is \( \dfrac{x_0}{2} \) with angular frequency \( 2\omega \).
2. \( v=\dfrac{dx}{dt}=-x_0\omega\sin2\omega t \).
3. At the mean position \( \cos2\omega t=0 \), so \( \sin2\omega t=\pm1 \) and the speed is \( x_0\omega \).

Teacher's Note:
1. Use \( \cos^{2}\theta=\dfrac{1+\cos2\theta}{2} \) to see that the oscillation is about \( x_0/2 \), not about zero.
2. The maximum speed is amplitude × angular frequency \( =\dfrac{x_0}{2}\times2\omega \).

Answer: (c) simple harmonic with a period of 0.1 s

Explanation:
1. Use \( \sin\theta\cos\theta=\dfrac{1}{2}\sin2\theta \): \( y=0.1\sin(20\pi t+3\pi) \).
2. This is a single sine function of time, so the motion is SHM with \( \omega=20\pi \) rad/s.
3. \( T=\dfrac{2\pi}{20\pi}=0.1 \) s.

Teacher's Note:
1. A product of a sine and a cosine of the same angle can be written as one sine of double the angle.
2. The period is found from the coefficient of t inside the sine.

Answer: (d) 2f

Explanation:
1. For springs attached on the two sides the effective force constant is \( K_1+K_2 \), so \( f=\dfrac{1}{2\pi}\sqrt{\dfrac{K_1+K_2}{m}} \).
2. When both become 4 times, \( K_1+K_2 \) becomes 4 times.
3. \( f'=\dfrac{1}{2\pi}\sqrt{\dfrac{4(K_1+K_2)}{m}}=2f \).

Teacher's Note:
1. Springs on opposite sides of a mass behave like springs in parallel: the force constants add.
2. Frequency is proportional to the square root of the force constant.

Answer: (c) \( k_1=\dfrac{k(\eta+1)}{\eta} \), \( k_2=k(\eta+1) \)

Explanation:
1. Force constant is inversely proportional to the length: \( kL=k_1l_1=k_2l_2 \).
2. With \( L=l_1+l_2=(\eta+1)l_2 \): \( k_2=\dfrac{kL}{l_2}=k(\eta+1) \).
3. \( l_1=\dfrac{\eta L}{\eta+1} \), so \( k_1=\dfrac{kL}{l_1}=\dfrac{k(\eta+1)}{\eta} \).

Teacher's Note:
1. A shorter piece of the same spring is stiffer, so its force constant is larger.
2. The product (force constant × length) is the same for every piece of the spring.

Answer: (b) 10 J

Explanation:
1. \( U=\dfrac{1}{2}kx^{2} \) with \( x=\dfrac{A}{2} \): \( U=\dfrac{1}{4}\left(\dfrac{1}{2}kA^{2}\right)=\dfrac{E}{4} \).
2. \( E=4U=4\times2.5=10 \) J.

Teacher's Note:
1. Potential energy varies as the square of the displacement, so half the amplitude gives one quarter of the total energy.
2. Total energy equals the maximum potential energy.

Answer: (b) \( \dfrac{1}{15}\ln3 \)

Explanation:
1. 15 oscillations take 15 s since T = 1 s.
2. \( 15=45\,e^{-15\gamma} \), so \( e^{15\gamma}=3 \).
3. \( \gamma=\dfrac{\ln3}{15} \) s⁻¹.

Teacher's Note:
1. The number of oscillations multiplied by the time period gives the elapsed time.
2. Take natural logarithms on both sides to remove the exponential.

Answer: (b) y = 5 cos 10t

Explanation:
1. Equilibrium extension \( =\dfrac{mg}{k}=\dfrac{20\times10}{2000}=10 \) cm. The mass is released from rest at 15 cm, so the amplitude is \( 15-10=5 \) cm.
2. \( \omega=\sqrt{\dfrac{k}{m}}=\sqrt{\dfrac{2000}{20}}=10 \) rad/s.
3. The mass starts at the lowest point, so \( y=5\cos10t \).

Teacher's Note:
1. Measure the displacement from the equilibrium position, not from the natural length of the spring.
2. A particle released from rest at an extreme position has a cosine equation of motion.

Answer: (a) \( T_1<T_2 \)

Explanation:
1. With \( x=A\sin\omega t \), x = A/2 at \( \omega t=\dfrac{\pi}{6} \), so \( T_1=\dfrac{T}{12} \).
2. x = A at \( \omega t=\dfrac{\pi}{2} \), so \( T_2=\dfrac{T}{4}-\dfrac{T}{12}=\dfrac{T}{6} \).
3. \( T_1<T_2 \) (in fact \( T_2=2T_1 \)).

Teacher's Note:
1. The particle moves fastest near the mean position and slowest near the extremes.
2. So it needs more time to cover the second half of the distance than the first half.

Answer: (d) \( \dfrac{9}{4} \)

Explanation:
1. \( f=\dfrac{1}{2\pi}\sqrt{\dfrac{g}{l}} \), so \( l\propto\dfrac{1}{f^{2}} \).
2. \( \dfrac{l_1}{l_2}=\dfrac{f_2^{2}}{f_1^{2}}=\dfrac{3^{2}}{2^{2}}=\dfrac{9}{4} \).

Teacher's Note:
1. The pendulum with the smaller frequency is the longer one.
2. Frequency varies inversely as the square root of the length.

Answer: (d) \( \sqrt2+1 \)

Explanation:
1. \( x=A\sin\dfrac{2\pi t}{8}=A\sin\dfrac{\pi t}{4} \). In 1 s the distance is \( A\sin45^\circ=\dfrac{A}{\sqrt2} \).
2. At t = 2 s, \( x=A\sin90^\circ=A \), so the distance in the 2nd second is \( A-\dfrac{A}{\sqrt2} \).
3. Ratio \( =\dfrac{1/\sqrt2}{1-1/\sqrt2}=\dfrac{1}{\sqrt2-1}=\sqrt2+1 \).

Teacher's Note:
1. The particle moves fastest near the mean position, so more distance is covered in the first second.
2. The ratio must therefore be greater than 1.

Answer: (a) 1 : 2

Explanation:
1. Parallel: \( k_p=2k \), so \( f_p=\dfrac{1}{2\pi}\sqrt{\dfrac{2k}{m}} \).
2. Series: \( k_s=\dfrac{k}{2} \), so \( f_s=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{2m}} \).
3. \( \dfrac{f_s}{f_p}=\sqrt{\dfrac{k/2}{2k}}=\dfrac{1}{2} \).

Teacher's Note:
1. Parallel springs add their force constants and series springs add the reciprocals.
2. The parallel arrangement is stiffer, so it has the higher frequency.

Answer: (b) \( 2\pi\sqrt{\dfrac{1}{\beta}} \)

Explanation:
1. Let \( X=x-2 \) (displacement from the equilibrium position x = 2). Then \( \dfrac{d^{2}X}{dt^{2}}=-\beta X \).
2. This is SHM with \( \omega^{2}=\beta \).
3. \( T=\dfrac{2\pi}{\omega}=2\pi\sqrt{\dfrac{1}{\beta}} \).

Teacher's Note:
1. Shifting the origin to the equilibrium position does not change the time period.
2. Only the coefficient of the displacement term decides ω.

Answer: (b) \( \dfrac{2\pi}{15} \) s

Explanation:
1. Without the wall, \( T_0=2\pi\sqrt{\dfrac{0.1}{10}}=\dfrac{\pi}{5} \) s for amplitude 6°.
2. From the 6° extreme to the vertical takes \( \dfrac{T_0}{4} \), and from the vertical to 3° (half the amplitude) takes \( \dfrac{T_0}{12} \), so one way takes \( \dfrac{T_0}{3} \).
3. The return trip takes the same time, so the period is \( \dfrac{2T_0}{3}=\dfrac{2\pi}{15} \) s.

Teacher's Note:
1. For small angles the swing is SHM, and a wall only cuts the motion short on one side.
2. The time from the centre to half the amplitude is one-twelfth of the period.

Answer: (b) \( \dfrac{1}{4}m\omega^{2}A^{2} \)

Explanation:
1. \( K=\dfrac{1}{2}m\omega^{2}A^{2}\cos^{2}\omega t \).
2. The average of \( \cos^{2}\omega t \) over a period is \( \dfrac{1}{2} \).
3. \( \langle K\rangle=\dfrac{1}{4}m\omega^{2}A^{2} \), which is half of the total energy.

Teacher's Note:
1. The average kinetic energy and the average potential energy over a period are each half of the total energy.
2. The average of \( \cos^{2} \) or \( \sin^{2} \) over a full period is one-half.

Answer: (b) as x increases, k increases

Explanation:
1. For SHM the force must be \( F=-Kx \) with a constant K.
2. Equating, \( -k\sqrt{x}=-Kx \), so \( k=K\sqrt{x} \).
3. Hence k must increase with x (as \( \sqrt{x} \)).

Teacher's Note:
1. SHM needs a restoring force proportional to the displacement.
2. If the formula shows a different power of x, the factor in front must vary to compensate.

Answer: (c) \( \dfrac{\sqrt3}{2}T \)

Explanation:
1. In the accelerating lift \( g_{eff}=g+\dfrac{g}{3}=\dfrac{4g}{3} \).
2. \( T'=2\pi\sqrt{\dfrac{l}{g_{eff}}}=T\sqrt{\dfrac{3}{4}} \).
3. \( T'=\dfrac{\sqrt3}{2}T \).

Teacher's Note:
1. In a lift accelerating upward the effective gravity increases, so the period decreases.
2. In a freely falling lift \( g_{eff}=0 \) and the pendulum would not oscillate.

Answer: (c) \( \dfrac{MgA^{2}}{2L} \)

Explanation:
1. At the extreme position the bob is raised by \( h=L(1-\cos\theta_0)\approx\dfrac{L\theta_0^{2}}{2} \), where \( \theta_0=\dfrac{A}{L} \).
2. \( h=\dfrac{A^{2}}{2L} \).
3. \( U=Mgh=\dfrac{MgA^{2}}{2L} \).

Teacher's Note:
1. Use the small-angle approximation \( 1-\cos\theta\approx\theta^{2}/2 \).
2. The same energy appears as kinetic energy \( \dfrac{1}{2}Mv^{2} \) at the lowest point.

Answer: (d) 5.65 m/s and \( 5.33\times10^{2} \) m/s²

Explanation:
1. \( \omega=2\pi f=2\pi\times15=30\pi\approx94.2 \) rad/s.
2. \( v_{max}=\omega A=94.2\times0.06\approx5.65 \) m/s.
3. \( a_{max}=\omega^{2}A=(94.2)^{2}\times0.06\approx5.33\times10^{2} \) m/s².

Teacher's Note:
1. Maximum velocity is \( \omega A \) (at the mean position) and maximum acceleration is \( \omega^{2}A \) (at the extremes).
2. Always change frequency into angular frequency by multiplying with 2π.

Answer: (d) 0.5 s

Explanation:
1. The speed is maximum when the object passes through the mean position, x = 0.
2. \( \cos\pi t=0 \) first at \( \pi t=\dfrac{\pi}{2} \).
3. \( t=0.5 \) s.

Teacher's Note:
1. Speed is maximum at the mean position, where the displacement is zero.
2. The first time the cosine becomes zero is at a phase of π/2.