JEE Physics Oscillations MCQs Set 20

Answer: (b) \( \pi/3 \)

Explanation:
1. Let \( x=A\sin\phi \). Then \( \sin\phi=\sqrt3/2 \), so \( \phi=\pi/3 \) (moving towards +A) or \( 2\pi/3 \) (moving back).
2. Phase difference = \( 2\pi/3-\pi/3=\pi/3 \).

Teacher's Note:
1. Two phases give the same displacement, one on the way out and one on the way back. Their difference is twice the angle measured from the extreme position.

Answer: (d) \( 7T/12 \)

Explanation:
1. Path in one oscillation = 4A, so the required path = \( \tfrac58\times4A=2.5A \).
2. 2A (mean to extreme and back to the mean) takes T/2. The remaining 0.5A from the mean takes \( T/12 \) (since \( \sin30^\circ=\tfrac12 \)).
3. Total time = \( T/2+T/12=7T/12 \).

Teacher's Note:
1. Break the path into whole half-oscillations plus a part. Moving from the mean position to A/2 takes one twelfth of the period.

Answer: (c) \( \dfrac{\pi}{2\sqrt5}\ \text{s} \)

Explanation:
1. Divide by 4: \( \ddot x+80x=0 \), so \( \omega^2=80 \), \( \omega=4\sqrt5 \).
2. \( T=\dfrac{2\pi}{\omega}=\dfrac{2\pi}{4\sqrt5}=\dfrac{\pi}{2\sqrt5}\ \text{s} \).

Teacher's Note:
1. Write the equation in the standard form x double dot plus omega squared x equals zero. Do not forget to divide by the coefficient of the second derivative.

Answer: (a) \( \pi/2\ \text{s} \)

Explanation:
1. Amplitude A = 10 cm and maximum speed \( v_{max}=A\omega=40 \) cm/s, so \( \omega=4\ \text{s}^{-1} \).
2. \( T=2\pi/\omega=\pi/2\ \text{s} \).

Teacher's Note:
1. The intercepts of the v-x ellipse give amplitude and maximum speed. Their ratio is the angular frequency.

Answer: (d) \( 4\sin\left(\dfrac{\pi}{6}t+\dfrac{\pi}{6}\right) \)

Explanation:
1. Successive zero crossings are half a period apart, so \( T/2=11-5=6 \) s, \( T=12 \) s and \( \omega=\pi/6 \).
2. Amplitude is 4 cm. At t = 0, \( 2=4\sin\phi\Rightarrow\phi=\pi/6 \).
3. Check: at t = 5, \( \tfrac{5\pi}{6}+\tfrac\pi6=\pi \), which is a downward crossing.

Teacher's Note:
1. Use the zero crossings for the period and the starting value for the phase. A quick check of a crossing time confirms the phase.

Answer: (d) 31.4 cm/s

Explanation:
1. \( v=\dfrac{dx}{dt}=-20\pi\sin\left(2\pi t+\dfrac\pi2\right)=-20\pi\cos2\pi t \).
2. At t = 1/6: \( \cos60^\circ=0.5 \), so \( |v|=10\pi=31.4 \) cm/s.

Teacher's Note:
1. Differentiate, then substitute the time. Keep the phase inside the trigonometric function.

Answer: (c) \( \pi\sqrt{\dfrac m\beta} \)

Explanation:
1. \( F=-\dfrac{dU}{dx}=-4\beta x \), so the force constant is \( k=4\beta \).
2. \( T=2\pi\sqrt{m/k}=2\pi\sqrt{\dfrac{m}{4\beta}}=\pi\sqrt{\dfrac m\beta} \).

Teacher's Note:
1. The constant alpha does not affect the force. Compare U with one half k x squared to read off k.

Answer: (a) 1 s, 0.628 m/s, 4 m/s²

Explanation:
1. \( T=2\pi/\omega=1\ \text{s} \).
2. \( v_{max}=A\omega=0.1\times6.28=0.628\ \text{m/s} \).
3. \( a_{max}=\omega^2A=(6.28)^2(0.1)\approx4\ \text{m/s}^2 \).

Teacher's Note:
1. Maximum speed is amplitude times omega and maximum acceleration is amplitude times omega squared. Convert 10 cm to metres.

Answer: (a) 50.2 cm/s

Explanation:
1. \( v=\omega\sqrt{A^2-x^2}=6.28\sqrt{100-36} \).
2. \( =6.28\times8=50.2\ \text{cm/s} \).

Teacher's Note:
1. The 6-8-10 triangle makes the root exact. Speed falls from maximum at the mean position to zero at the extreme.

Answer: (a) -54.4 cm/s

Explanation:
1. Starting from the extreme: \( x=A\cos\omega t \), so \( v=-A\omega\sin\omega t \).
2. At t = 1/6 s, \( \omega t=2\pi/6=60^\circ \): \( v=-62.8\times\sin60^\circ=-54.4\ \text{cm/s} \).

Teacher's Note:
1. Choose cosine when the motion starts at the extreme. The negative sign shows the particle is moving towards the mean position.

Answer: (a) \( \theta=\dfrac\pi{10}\cos40\pi t \)

Explanation:
1. \( \omega=2\pi/0.05=40\pi\ \text{rad/s} \).
2. At t = 0 the displacement equals the amplitude, so the cosine form is used: \( \theta=\dfrac\pi{10}\cos40\pi t \).

Teacher's Note:
1. The same equations hold for angular SHM. Starting at the extreme means cosine with zero phase.

Answer: (a) 2 m/s and 4 m/s²

Explanation:
1. \( \omega^2=4\Rightarrow\omega=2\ \text{rad/s} \); amplitude A = 1 m.
2. \( v_{max}=A\omega=2\ \text{m/s} \), \( a_{max}=\omega^2A=4\ \text{m/s}^2 \).

Teacher's Note:
1. Read omega squared as the coefficient in the acceleration equation. Use amplitude to get the maxima.

Answer: (a) 195°

Explanation:
1. Phase of \( Y_1 \) at 0.5 s = \( 5\pi+\pi/3 \); phase of \( Y_2 \) = \( 4\pi+\pi/4 \).
2. Difference = \( \pi+\pi/3-\pi/4=\pi+\pi/12=195^\circ \).

Teacher's Note:
1. Phase is the whole argument of the sine, not only the constant term. Subtract the two arguments at the given time.

Answer: (d) \( a/\sqrt2 \)

Explanation:
1. \( y=a\sin\omega t=a\sin\left(\dfrac{2\pi}{T}\cdot\dfrac T8\right)=a\sin45^\circ=\dfrac a{\sqrt2} \).

Teacher's Note:
1. Time measured from the mean position uses the sine form. One eighth of a period is 45 degrees of phase.

Answer: (c) \( \pi \)

Explanation:
1. Acceleration \( a=-\omega^2x \) is always opposite to displacement, so they are out of phase by \( \pi \).

Teacher's Note:
1. Velocity leads displacement by pi by two and acceleration leads velocity by another pi by two. Together that is pi.

Answer: (d) simple harmonic

Explanation:
1. Write \( a=-100(x-0.5) \): the acceleration is proportional to the displacement from x = 0.5 m and directed towards it.
2. This is the condition for SHM, with the equilibrium shifted to x = 0.5 m.

Teacher's Note:
1. A constant term only shifts the centre of oscillation. The restoring nature is unchanged.

Answer: (a) 2.0 m

Explanation:
1. The maximum acceleration is \( \omega^2A=35.28 \) with \( \omega=4.2 \).
2. \( A=\dfrac{35.28}{(4.2)^2}=\dfrac{35.28}{17.64}=2.0\ \text{m} \).

Teacher's Note:
1. The coefficient of the cosine is the maximum acceleration, not the amplitude. Divide by omega squared.

Answer: (d) Infinite

Explanation:
1. The total mechanical energy of an ideal oscillator is constant.
2. A constant does not oscillate, so its period is infinite.

Teacher's Note:
1. Kinetic and potential energy each oscillate with half the period, T/2. Their sum does not change.

Answer: (d) \( \dfrac{2\pi\sqrt2}{5}\ \text{s} \)

Explanation:
1. \( U=\tfrac12kx^2\Rightarrow k=\dfrac{2(1.0)}{(0.2)^2}=50\ \text{N/m} \).
2. \( T=2\pi\sqrt{m/k}=2\pi\sqrt{4/50}=2\pi\sqrt{\dfrac{2}{25}}=\dfrac{2\pi\sqrt2}{5}\ \text{s} \).

Teacher's Note:
1. Read the force constant from the potential energy curve. Then use T equals two pi root m by k.

Answer: (a) 3.5 kg

Explanation:
1. \( T\propto\sqrt m \): \( \dfrac{5^2}{4^2}=\dfrac{m+2}{m} \).
2. \( 25m=16m+32\Rightarrow m=\dfrac{32}{9}\approx3.5\ \text{kg} \).

Teacher's Note:
1. Take the ratio of the periods to eliminate k. Square before cross-multiplying.

Answer: (b) \( \dfrac{160}{\pi^2}\ \text{kg} \)

Explanation:
1. Force constant \( k=10/1=10\ \text{N/m} \); angular frequency \( \omega=2\pi/8=\pi/4 \).
2. \( m=\dfrac{k}{\omega^2}=\dfrac{10}{\pi^2/16}=\dfrac{160}{\pi^2}\ \text{kg} \).

Teacher's Note:
1. Slope of the F-x line gives k and the x-t graph gives omega. Mass follows from omega squared equals k by m.

Answer: (c) \( 8.0\times10^{-3}\ \text{J} \)

Explanation:
1. \( K_{max}=\tfrac12kA^2=\tfrac12(10)(0.04)^2=8\times10^{-3}\ \text{J} \).

Teacher's Note:
1. Maximum kinetic energy equals the total energy, half k A squared. Convert the amplitude to metres.

Answer: (b) 100 J

Explanation:
1. Energy exchanged in the oscillation = \( \tfrac12kA^2=\tfrac12(2\times10^6)(10^{-4})=100\ \text{J} \).
2. The total of 160 J includes a constant potential energy of 60 J at the mean position, so the maximum kinetic energy is 160 - 60 = 100 J.

Teacher's Note:
1. Total energy depends on where potential energy is taken to be zero. The kinetic energy swing is always half k A squared.

Answer: (c) 7 m

Explanation:
1. \( F=-dU/dx=20-10x=0\Rightarrow x_0=2\ \text{m} \) (equilibrium).
2. Amplitude = 2 - (-3) = 5 m, so the maximum x = 2 + 5 = 7 m.

Teacher's Note:
1. Find the equilibrium position first. The particle swings equally far on both sides of it.

Answer: (d) 20 s

Explanation:
1. Potential energy is maximum at an extreme position and minimum (zero) at the mean position.
2. That takes a quarter of the period: \( T/4=5 \) s, so T = 20 s.

Teacher's Note:
1. Extreme to mean is a quarter period. Do not use half the period here.