JEE Physics Oscillations MCQs Set 17

Answer: (b) 2

Explanation:
1. The separation is a SHM of amplitude \( 2a\sin(\Delta\phi/2) \).
2. \( 2a\sin(\Delta\phi/2) = a\sqrt2 \Rightarrow \Delta\phi/2 = \pi/4 \), so \( \Delta\phi = \pi/2 \) and N = 2.

Teacher's Note:
1. The relative displacement of two equal SHMs is itself SHM.
2. Use the phasor difference.

Answer: (a) 2

Explanation:
1. The acceleration \( a-bx = -b\left(x-\dfrac ab\right) \) is that of SHM about \( x_0 = a/b \).
2. The car starts at rest at x = 0, so the amplitude is a/b and the car comes to rest again at \( x = 2a/b \). Hence N = 2.

Teacher's Note:
1. Rewrite the acceleration about the point where it vanishes.
2. A body starting from rest at one extreme comes to rest at the opposite extreme.

Answer: (d) 15

Explanation:
1. Angular frequency \( \omega = n \). Speed before the blow: \( v = n\sqrt{a^2-3a^2/4} = na/2 \).
2. After the blow \( v' = \tfrac{na}{2}+na = \tfrac{3na}{2} \). New amplitude: \( A'^2 = x^2+v'^2/n^2 = \tfrac34a^2+\tfrac94a^2 = 3a^2 \).
3. \( A' = \sqrt3a = 15\sqrt3 \) cm, so k = 15.

Teacher's Note:
1. The blow changes the velocity but not the position.
2. Use A squared equals x squared plus v squared over omega squared.

Answer: (c) 2

Explanation:
1. Same period and same phase: \( x_1-x_2 = (A_1-A_2)\cos\omega t = 12\cos\omega t \).
2. \( 12\cos\omega t = 6 \Rightarrow \omega t = \pi/3 \), so \( t = T/6 = 2 \) s.

Teacher's Note:
1. Both start from positive extremes, so the separation oscillates as a cosine.
2. 60 degrees is one sixth of the period.

Answer: (b) 16

Explanation:
1. Acceleration along each incline \( = g\sin30^\circ = 5 \) m/s\(^2\). Length of the incline \( = 0.20/\sin30^\circ = 0.40 \) m.
2. Time from A to B: \( 0.40 = \tfrac12(5)t^2 \Rightarrow t = 0.4 \) s.
3. One full period consists of four such passages: T = 1.6 s = 16 x 10 to the minus 1 s.

Teacher's Note:
1. The motion is a uniformly accelerated slide on each side, not SHM.
2. A full period is down, up, down and up again.

Answer: (a) 3

Explanation:
1. For a displacement along C, the effective stiffness of a spring with axis at angle \( \theta \) to the displacement is \( k\cos^2\theta \).
2. C: \( 2k\cos^20^\circ = 2k \). A and B each make 45 degrees: \( k\cos^245^\circ = k/2 \) each.
3. \( k_{eff} = 2k+\tfrac k2+\tfrac k2 = 3k \), so \( \omega = \sqrt{3k/m} \) and N = 3.

Teacher's Note:
1. Only the component of each spring's force along the displacement matters.
2. Springs with axes at an angle contribute k times cos squared.

Answer: (d) 3

Explanation:
1. Speed of the particle before impact: \( v = \sqrt{2g\times4.5mg/K} = 3g\sqrt{m/K} \). After sticking (momentum conservation) the speed of \( 3m \) is \( g\sqrt{m/K} \).
2. New equilibrium is lower by \( mg/K \), so the combined mass starts \( mg/K \) above the new equilibrium, moving down, with \( \omega = \sqrt{K/3m} \). Amplitude \( A = \sqrt{(mg/K)^2+3g^2m^2/K^2} = 2mg/K \).
3. The initial displacement is -A/2 (above equilibrium) moving down; the lowest point is reached after a phase of 30 + 90 = 120 degrees, i.e. \( t = \dfrac{2\pi/3}{\omega} = 2\pi\sqrt{\dfrac{m}{3K}} \). So a = 3.

Teacher's Note:
1. Find the speed after the perfectly inelastic collision from momentum conservation.
2. Measure the starting displacement from the new equilibrium position.

Answer: (c) 15

Explanation:
1. Momentum conservation at the mean position: \( 4v = 9v' \), with \( v = A\omega = 0.1\times5 = 0.5 \) m/s, so \( v' = 2/9 \) m/s.
2. New \( \omega' = \sqrt{100/9} = 10/3 \) rad/s, so \( A' = v'/\omega' = \dfrac{2/9}{10/3} = \dfrac1{15} \) m.
3. So n = 15.

Teacher's Note:
1. At the mean position the amplitude is just speed divided by omega.
2. The placement is a perfectly inelastic process in the horizontal direction.

Answer: (b) 8

Explanation:
1. The vehicle's acceleration down the plane is \( g\sin\alpha \). The effective gravity relative to the vehicle is \( g_{eff} = g\cos\alpha = g/2 \) (perpendicular to the plane).
2. \( T = 2\pi\sqrt{L/(g/2)} = 2\pi\sqrt{2L/g} = \pi\sqrt{8L/g} \), so X = 8.

Teacher's Note:
1. Subtract the vehicle's acceleration from g as a vector.
2. Only the component of g perpendicular to the incline remains.

Answer: (a) 15

Explanation:
1. Maximum KE equals the gain in potential energy: \( mg\ell(1-\cos\theta_0)\approx\tfrac12mg\ell\theta_0^2 = 15\times10^{-3} \).
2. \( \ell = \dfrac{2\times0.015}{0.2\times10\times0.01} = 1.5 \) m \( = \dfrac{15}{g} \), so n = 15.

Teacher's Note:
1. Energy conservation links the turning angle and the maximum kinetic energy.
2. Use the small-angle form of 1 minus cos.

Answer: (d) 2

Explanation:
1. In water the net downward force is reduced by buoyancy: \( g_{eff} = g\left(1-\dfrac{\rho_w}{\rho}\right) = g\left(1-\dfrac34\right) = \dfrac g4 \).
2. \( T\propto1/\sqrt{g_{eff}} \Rightarrow t = 2t_0 \).

Teacher's Note:
1. Buoyancy reduces the effective weight per unit mass.
2. The period doubles when g_eff falls to a quarter.

Answer: (c) 8

Explanation:
1. Mass of the sphere \( m = \tfrac12\rho\times\tfrac43\pi R^3 = \tfrac23\pi\rho R^3 \). Extra buoyant force for a small push x: \( \rho g\,\pi R^2\,x \).
2. \( \omega^2 = \dfrac{\rho g\pi R^2}{\tfrac23\pi\rho R^3} = \dfrac{3g}{2R} \), so \( f = \dfrac{1}{2\pi}\sqrt{\dfrac{3g}{2R}} \).
3. With \( \pi = \sqrt g \): \( f = \dfrac{1}{2\sqrt g}\sqrt{\dfrac{3g}{2R}} = \sqrt{\dfrac{3}{8R}} \), so n = 8.

Teacher's Note:
1. A body floating exactly half submerged has density equal to half the liquid.
2. The surface area at the waterline is pi R squared for a small displacement.

Answer: (b) 75

Explanation:
1. Moment of inertia about the hinge: \( I = m\ell^2/3 \).
2. Restoring torque for a small angle \( \theta \): gravity \( mg\dfrac\ell2\theta \) plus springs \( K\ell\theta\times\ell = K\ell^2\theta \).
3. \( \omega^2 = \dfrac{mg\ell/2+K\ell^2}{m\ell^2/3} = \dfrac{3g}{2\ell}+\dfrac{3K}{m} = \dfrac{7.5g}{\ell} = \dfrac{75}{\ell} \), so n = 75.

Teacher's Note:
1. Both gravity and the springs give restoring torques about the hinge.
2. Both springs act on the same side as torque, so K = K/2 + K/2.

Answer: (b) (A), (B), (C), (D) and (E)

Explanation:
1. Losing contact at the top: \( \omega^2A = g \Rightarrow \omega = \sqrt{10/0.4} = 5 \) rad/s, so \( T = 2\pi/5 \) s (A).
2. Normal reaction \( N = m(g\pm\omega^2x) \). At the lowest point \( \omega^2A = g \) upwards, so \( N = 2mg \) (B). Halfway up the acceleration is g/2 downwards: \( N = 0.5mg \) (C); halfway down g/2 upwards: \( N = 1.5mg \) (D). At the mean position the acceleration is zero: \( N = mg \) (E).
3. All five statements are correct.

Teacher's Note:
1. The reaction depends on the acceleration of the plank.
2. At the top the acceleration equals g downwards, so the reaction is zero.

Answer: (d) (A), (B), (C) and (D)

Explanation:
1. At \( y = A/2 \): \( v = v_{max}\cos30^\circ = 0.866v_{max}>v_{max}/2 \) (A). For \( v = v_{max}/2 \): \( \cos\omega t = 1/2 \), \( \omega t = 60^\circ \), \( y = 0.866A>A/2 \) (B).
2. At \( t = T/8 \), \( \omega t = 45^\circ \), \( y = 0.707A>A/2 \) (C); \( y = A/2 \) occurs at \( t = T/12

Teacher's Note:
1. Use the reference circle with the angles 30, 45 and 60 degrees.
2. In the first quarter-cycle the speed falls as the displacement rises.

Answer: (c) (B), (C) and (D)

Explanation:
1. \( U = 5x^2-20x \), so \( F = -dU/dx = -10(x-2) \): a restoring force about x = 2 m with \( k = 10 \) N/m, not a constant force (A false, C true).
2. The speed is maximum at the equilibrium position x = 2 m (B true).
3. \( T = 2\pi\sqrt{m/k} = 2\pi\sqrt{0.1/10} = \pi/5 \) s (D true).

Teacher's Note:
1. The force is minus the slope of the potential energy curve.
2. A linear restoring force means SHM about the point where the force vanishes.

Answer: (a) \( 2\sqrt{\dfrac{\ell}{g}}\left[\dfrac\pi2+\sin^{-1}\dfrac\alpha\beta\right] \)

Explanation:
1. The angle varies as \( \theta = \beta\cos\omega t \) with \( \omega = \sqrt{g/\ell} \), until the thread strikes the wall at \( \theta = -\alpha \).
2. This happens when \( \cos\omega t = -\alpha/\beta \), i.e. at \( t_1 = \dfrac1\omega\cos^{-1}(-\alpha/\beta) = \dfrac1\omega\left[\dfrac\pi2+\sin^{-1}\dfrac\alpha\beta\right] \).
3. After the elastic bounce the ball retraces its path, so the period is \( 2t_1 \).

Teacher's Note:
1. The wall cuts off the swing at the angle alpha on the far side.
2. The period is twice the time from the release to the wall.

Answer: (d) (B) and (C)

Explanation:
1. \( \sin\omega t+\cos\omega t = \sqrt2\sin(\omega t+\pi/4) \), so the amplitude is \( \sqrt2 \) m (B true, A false).
2. At t = 0, y = sin 0 + cos 0 = 1 m, so time is measured from y = 1 m (C true, D false).

Teacher's Note:
1. Combine the two terms into one sinusoid.
2. Substitute t = 0 to find the starting position.

Answer: (c) (A), (C) and (D)

Explanation:
1. Both coordinates are proportional to \( \cos\omega t \): \( y = 2x \), a straight line (A).
2. The motion along that line is \( \sqrt5A\cos\omega t \), which is periodic (C) and simple harmonic (D). It is not an ellipse.

Teacher's Note:
1. Both components have the same phase, so the path is a straight line.
2. Along the line the displacement is a pure cosine.

Answer: (b) (A) and (C)

Explanation:
1. Phasors at 0, 45 and 90 degrees: resultant along the middle one (45 degrees) with amplitude \( a+2a\cos45^\circ = (1+\sqrt2)a \) (A true). Its phase relative to the first is 45 degrees, not 90 (B false).
2. Energy is proportional to amplitude squared: \( (1+\sqrt2)^2 = 3+2\sqrt2 \) (C true). The sum is a single sinusoid, so it is SHM (D false).

Teacher's Note:
1. Use symmetry: the resultant lies along the middle phasor.
2. Energy of an oscillator scales with the square of its amplitude.

Answer: (a) \( 10\sqrt2 \) rad/s

Explanation:
1. \( \omega = \sqrt{k/m} = \sqrt{400/2} = 10\sqrt2 \) rad/s.

Teacher's Note:
1. The angular frequency depends only on k and m.
2. The motion of the elevator at constant speed does not change it.

Answer: (d) 2.5 cm

Explanation:
1. Before t = 0 the spring is stretched by \( x_1 = \dfrac{m(g+a)}{k} = \dfrac{2\times15}{400} = 7.5 \) cm.
2. After the acceleration ceases the equilibrium extension is \( mg/k = 5 \) cm, so the amplitude is \( 7.5-5 = 2.5 \) cm.

Teacher's Note:
1. The block starts at rest at the old equilibrium position.
2. Amplitude is the distance from the new equilibrium.

Answer: (c) \( 3\pi/2 \) rad

Explanation:
1. At t = 0 the block is at the old equilibrium, which is 2.5 cm below the new equilibrium (the spring is more stretched), i.e. at the negative extreme.
2. The positive extreme has phase \( \pi/2 \), so the negative extreme has phase \( \pi/2+\pi = 3\pi/2 \).

Teacher's Note:
1. Decide whether the starting position is above or below the new equilibrium.
2. Opposite extremes differ by pi in phase.

Answer: (b) \( 2\pi\sqrt{\dfrac{a}{3g}} \)

Explanation:
1. Let \( AP = x \). Net force towards B \( = \dfrac{2mg(a-x)}{a}-\dfrac{mgx}{a} = \dfrac{mg(2a-3x)}{a} \).
2. This is \( -\dfrac{3mg}{a}\left(x-\dfrac{2a}{3}\right) \): SHM about \( x = 2a/3 \) with \( k = 3mg/a \).
3. \( T = 2\pi\sqrt{m/k} = 2\pi\sqrt{\dfrac{a}{3g}} \).

Teacher's Note:
1. Find the net force and write it in the form minus k times the displacement from equilibrium.
2. The equilibrium is where the two forces balance.

Answer: (a) a/6

Explanation:
1. Equilibrium at \( x = 2a/3 \) (from A); the release point is the mid-point \( x = a/2 \), so the amplitude is \( 2a/3-a/2 = a/6 \).
2. The other extreme is at \( 2a/3+a/6 = 5a/6 \), which is at distance \( a-5a/6 = a/6 \) from B.

Teacher's Note:
1. The motion is symmetric about the equilibrium point.
2. Minimum distance from B is at the extreme nearest B.

Answer: (d) \( \dfrac{\sqrt{2ga}}{6} \)

Explanation:
1. When nearest to B the particle is at rest at distance \( a/6 \) from B. Only the force \( \dfrac{2mg(BP)}{a} \) towards B acts afterwards.
2. Work done over the distance \( a/6 \): \( W = \int_0^{a/6}\dfrac{2mg\,u}{a}du = \dfrac{mg}{a}\left(\dfrac a6\right)^2 = \dfrac{mga}{36} \).
3. \( \tfrac12mv^2 = \dfrac{mga}{36}\Rightarrow v = \sqrt{\dfrac{ga}{18}} = \dfrac{\sqrt{2ga}}{6} \).

Teacher's Note:
1. The force varies linearly with the distance from B, so use the work done by a linear force.
2. Starting at rest at the nearest point.

Answer: (b) \( \sqrt{\dfrac{3}{k}(m_1+m_2)}\;g\sin\theta \)

Explanation:
1. Let \( d = \dfrac{(m_1+m_2)g\sin\theta}{k} \). The blocks separate when the spring reaches its natural length (there the contact force vanishes), at a compression of \( 3d \) below the start.
2. Energy: \( \tfrac12(m_1+m_2)v^2 = \tfrac12k(3d)^2-(m_1+m_2)g\sin\theta\,(3d) = 4.5kd^2-3kd^2 = 1.5kd^2 \).
3. \( v^2 = \dfrac{3kd^2}{m_1+m_2} = \dfrac{3(m_1+m_2)g^2\sin^2\theta}{k} \), so \( v = \sqrt{\dfrac3k(m_1+m_2)}\;g\sin\theta \).

Teacher's Note:
1. The blocks separate when the spring force on m1 becomes zero, at the natural length.
2. Use energy conservation between the release point and the natural-length position.

Answer: (a) (A), (B) and (D)

Explanation:
1. \( x = \dfrac{A+B}{2}+\dfrac{B-A}{2}\cos2\omega t+\dfrac C2\sin2\omega t \): a sinusoid of angular frequency \( 2\omega \) about \( (A+B)/2 \), with amplitude \( \sqrt{\left(\tfrac{B-A}2\right)^2+\left(\tfrac C2\right)^2} \). It is SHM whenever this amplitude is non-zero, so (A) is true.
2. A = -B, C = 2B: amplitude \( \sqrt{B^2+B^2} = |B|\sqrt2 \) (B true). A = B, C = 2B: amplitude \( |C/2| = |B| \) (D true).
3. A = B and C = 0 gives a constant x = A: no oscillation, so (C) is false.

Teacher's Note:
1. Rewrite using double-angle identities.
2. The amplitude is zero (no motion) only if both coefficients vanish.