JEE Physics Oscillations MCQs Set 07

Answer: (b) 2 cm

Explanation:
1. Energy stored in the left spring is \( \tfrac12 (100)(0.04)^2 = 0.08 \) J.
2. This energy passes to the right spring: \( \tfrac12 (400)x^2 = 0.08 \), so \( x = 0.02 \) m = 2 cm.

Teacher's Note:
1. Energy is conserved on the smooth surface.
2. Stiffer spring gives smaller compression.

Answer: (a) 2.0 m

Explanation:
1. Compare with \( a = -\omega^2 A\cos\omega t \): \( \omega = 4.2 \) rad/s.
2. \( A = 35.28/(4.2)^2 = 35.28/17.64 = 2.0 \) m.

Teacher's Note:
1. Maximum acceleration is \( \omega^2 A \).
2. Divide by \( \omega^2 \).

Answer: (d) \( \sqrt{\dfrac{b}{a}} \)

Explanation:
1. Write the equation as \( \ddot{x} = -(b/a)x \), so \( \omega^2 = b/a \).
2. Maximum acceleration \( \omega^2 A \) divided by maximum velocity \( \omega A \) is \( \omega = \sqrt{b/a} \).

Teacher's Note:
1. \( \omega^2 \) is the coefficient of -x.
2. Ratio = \( \omega \).

Answer: (c) 7

Explanation:
1. \( T = 2\pi\sqrt{\ell/g} = 2\pi/\sqrt{10} = 2 \) s since \( \pi^2 = 10 \).
2. Kinetic energy is maximum at the mean position, which is reached at t = 0.5, 1.5, 2.5, ... s (every T/2 starting at T/4). Up to 6.7 s these are 0.5, 1.5, 2.5, 3.5, 4.5, 5.5, 6.5, which is 7 times.

Teacher's Note:
1. Release from an extreme position.
2. Mean position is crossed twice per period.

Answer: (d) \( y = 0.5\cos 5\pi t \)

Explanation:
1. \( \omega = 2\pi/T = 2\pi/0.4 = 5\pi \) rad/s.
2. \( y = 0.5\sin(5\pi t + \pi/2) = 0.5\cos 5\pi t \).

Teacher's Note:
1. Add the phase inside the sine.
2. sin(x + pi/2) = cos x.

Answer: (b) 1.25 s

Explanation:
1. Periods: \( T_1 = 2\pi\sqrt{0.25/\pi^2} = 1 \) s and \( T_2 = 2\pi\sqrt{1/\pi^2} = 2 \) s.
2. First collision at \( T_1/4 = 0.25 \) s at the mean position. For an elastic collision between equal masses the velocities are exchanged, so the long pendulum goes out and returns to the mean position after \( T_2/2 = 1 \) s. Second collision at 0.25 + 1 = 1.25 s.

Teacher's Note:
1. Equal masses exchange velocities.
2. The second bob needs half its period to return.

Answer: (b) 5 s

Explanation:
1. The time from the position x = 12 cm to the mean position is \( \tfrac{T}{2\pi}\sin^{-1}(12/15) \) and from the mean position to x = 9 cm is \( \tfrac{T}{2\pi}\sin^{-1}(9/15) \).
2. The angles are 53 degrees and 37 degrees, adding to 90 degrees, so the total time is \( T/4 = 5 \) s.

Teacher's Note:
1. Use x = A sin(omega t) from the mean position.
2. sin^-1(0.8) + sin^-1(0.6) = pi/2.

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

Explanation:
1. \( \omega = \sqrt{g/\ell} = \sqrt{10/0.1} = 10 \) rad/s. Take \( \theta = 2^\circ\cos 10t \).
2. \( \theta = -1^\circ \) gives \( \cos\phi = -\tfrac12 \). Moving towards the mean position from the left side means \( \phi = 4\pi/3 \), so \( t = 4\pi/30 = 2\pi/15 \) s.

Teacher's Note:
1. Left side is negative displacement.
2. Velocity must be towards the mean position.

Answer: (c) 1.0 s

Explanation:
1. \( \omega = 0.5\pi \), so \( T = 2\pi/\omega = 4 \) s.
2. Equilibrium to maximum displacement takes \( T/4 = 1 \) s.

Teacher's Note:
1. A quarter of a period.
2. Period from omega.

Answer: (d) \( \dfrac{T}{2\pi}\cos^{-1}\dfrac{4}{5} \)

Explanation:
1. Starting at the extreme, \( x = A\cos\omega t \). After a distance A/5 the displacement is \( 4A/5 \).
2. So \( \cos\omega t = 4/5 \) and \( t = \tfrac{T}{2\pi}\cos^{-1}\tfrac45 \).

Teacher's Note:
1. Displacement from the mean position is A - A/5.
2. Invert the cosine.

Answer: (b) 12 s

Explanation:
1. Amplitude decays as \( A e^{-bt/2m} \) with \( b = 0.230 \) kg/s and \( m = 2 \) kg.
2. For half amplitude, \( t = \tfrac{2m}{b}\ln 2 = \tfrac{4}{0.230}\times 0.693 \approx 12 \) s.

Teacher's Note:
1. Damping constant b.
2. Take natural log of 2.

Answer: (c) \( T < 2\pi\sqrt{\ell/g} \) and damping is larger than in air alone

Explanation:
1. The iron bob is attracted by the magnetic field of the coil, which adds a downward force, so effective g is greater and \( T < 2\pi\sqrt{\ell/g} \).
2. Also, the moving iron bob changes the magnetic flux and sets up eddy currents which oppose its motion, so damping is larger than in air alone.

Teacher's Note:
1. Extra downward pull shortens the period.
2. Eddy currents add damping.

Answer: (a) T/6

Explanation:
1. Take \( x = a\cos\omega t \). At \( x = a/2 \), \( \cos\omega t = 1/2 \), so \( \omega t = \pi/3 \).
2. \( t = \tfrac{\pi/3}{2\pi/T} = T/6 \).

Teacher's Note:
1. Start from the extreme position.
2. cos 60 degrees = 1/2.

Answer: (b) 0.729

Explanation:
1. Amplitude falls by the same factor in equal times: 0.9 for every 5 s.
2. In a further 10 s it falls by two more factors, so the amplitude is \( 0.9^3 = 0.729 \) of the original.

Teacher's Note:
1. Exponential decay.
2. Total time is 15 s, three intervals.

Answer: (a) \( \sqrt{45} \) and \( \tan^{-1}(1/2) \)

Explanation:
1. Write all in terms of cos and sin: \( y_2 = -4\sin\omega t \), \( y_3 = -2\cos\omega t \), \( y_4 = \sin\omega t \).
2. Sum \( = 6\cos\omega t - 3\sin\omega t = A\cos(\omega t + \phi) \) with \( A\cos\phi = 6 \), \( A\sin\phi = 3 \). So \( A = \sqrt{45} \) and \( \tan\phi = 1/2 \).

Teacher's Note:
1. Add components.
2. Compare with A cos(wt + phi).

Answer: (a) \( \dfrac{\pi A\sqrt{3}}{T} \)

Explanation:
1. \( v = \omega\sqrt{A^2 - x^2} = \tfrac{2\pi}{T}\sqrt{A^2 - A^2/4} \).
2. \( v = \tfrac{2\pi}{T}\cdot\tfrac{\sqrt3 A}{2} = \tfrac{\pi A\sqrt3}{T} \).

Teacher's Note:
1. Use the velocity formula.
2. Substitute x = A/2.

Answer: (d) 10 Hz

Explanation:
1. Frequency heard directly: \( 350\times 350/355 \). Frequency of the reflected sound: \( 350\times 350/345 \).
2. Beats = \( 122500(1/345 - 1/355) \approx 10 \) Hz.

Teacher's Note:
1. Source moves away from the person but towards the wall.
2. Subtract the two.

Answer: (c) zero

Explanation:
1. All move with the same velocity, so there is no relative motion between the sources and the observer.
2. There is no Doppler shift, both sources are heard at 100 Hz and the beat frequency is zero.

Teacher's Note:
1. Relative velocity is zero.
2. Same frequency, no beats.

Answer: (c) 5

Explanation:
1. Write \( x = A\sin(\omega t + \phi) \) with \( A\cos\phi = 3 \) and \( A\sin\phi = 4 \).
2. \( A = \sqrt{3^2 + 4^2} = 5 \).

Teacher's Note:
1. Combine into one sinusoid.
2. Amplitude = square root of sum of squares.

Answer: (c) a straight line

Explanation:
1. With the same phase, \( x = A\sin\omega t \) and \( y = A\sin\omega t \), so \( y = x \).
2. This is the equation of a straight line.

Teacher's Note:
1. Eliminate time.
2. Phase difference zero gives a line.

Answer: (a) air friction

Explanation:
1. The bob loses mechanical energy to the air by friction, so its amplitude falls and it finally stops.
2. Moment of inertia and weight are not dissipative.

Teacher's Note:
1. Air friction is the damping force.
2. Energy is dissipated.

Answer: (c) \( \pi \) s

Explanation:
1. \( v^2 = \omega^2(A^2 - x^2) \): \( 100 = \omega^2(A^2 - 16) \) and \( 64 = \omega^2(A^2 - 25) \).
2. Subtract: \( 36 = 9\omega^2 \), so \( \omega = 2 \) rad/s and \( T = 2\pi/2 = \pi \) s.

Teacher's Note:
1. Two equations, two unknowns.
2. Eliminate A.