JEE Physics Oscillations MCQs Set 11

Answer: (d) (b) and (d) only

Explanation:
1. \( 8\cos^250t=4(1+\cos100t) \), so \( x=4+(3\sin100t+4\cos100t) \).
2. The bracket is an SHM of amplitude \( \sqrt{9+16}=5 \) about the mean position x = 4, so the motion is SHM ((a), (c) false; (b) true).
3. The maximum value of x is \( 4+5=9 \) ((d) true).

Teacher's Note:
1. Use cos^2 (theta) = (1 + cos 2 theta)/2 to expose the constant term.
2. A constant term shifts the mean position without stopping the motion from being SHM.

Answer: (c) (b) and (c) only

Explanation:
1. Along the string: \( T-Mg\cos\theta=\frac{MV^2}{l} \) ((b) true; (a), (d) false because the bob has centripetal acceleration).
2. Perpendicular to the string: \( Mg\sin\theta=Ma_t\Rightarrow a_t=g\sin\theta \) ((c) true).

Teacher's Note:
1. Resolve forces along the string (radial) and perpendicular to it (tangential).
2. The tension equals Mg cos(theta) only when the speed is zero (at the extreme).

Answer: (b) (a) and (c) only

Explanation:
1. \( T_2=2\pi\sqrt{\frac{l}{g+a}}\lt T_1\lt T_3=2\pi\sqrt{\frac{l}{g-a}} \), so (a) is true and (b) is false.
2. \( \frac1{T_2^2}+\frac1{T_3^2}=\frac{(g+a)+(g-a)}{4\pi^2l}=\frac{2g}{4\pi^2l}=\frac{2}{T_1^2} \).
3. Hence \( T_1=\frac{\sqrt2\,T_2T_3}{\sqrt{T_2^2+T_3^2}} \) ((c) true, (d) false).

Teacher's Note:
1. Write 1/T^2 for each case and add the upward and downward cases so that the acceleration cancels.
2. Upward acceleration decreases the period, downward acceleration increases it.

Answer: (b) (c) and (d) only

Explanation:
1. \( x^2+(y-a)^2=a^2\sin^2\omega t+a^2\cos^2\omega t=a^2 \): a circle of radius a centred at (0, a) ((c) true).
2. The speed \( \sqrt{\dot x^2+\dot y^2}=a\omega \) is constant, so the distance travelled \( =a\omega t \) is proportional to time ((d) true).

Teacher's Note:
1. Eliminate time with sin^2 + cos^2 = 1.
2. Uniform circular motion has constant speed.

Answer: (a) (b) and (c) only

Explanation:
1. The two motions are in phase, so \( \frac{y}{x}=\frac34 \): a straight line.
2. Displacement along the line is \( r=\sqrt{x^2+y^2}=5\sin100\pi t \), an SHM of amplitude 5.

Teacher's Note:
1. Equal frequencies and zero phase difference give a straight-line path.
2. The resultant amplitude is sqrt(4^2 + 3^2) = 5.

Answer: (d) (a) and (c) only

Explanation:
1. Undamped SHM conserves total energy, which equals the maximum KE, so the average is that value ((a) true, (b) false).
2. \( v_{rms}=\frac{v_{max}}{\sqrt2} \) ((c) true).
3. The mean velocity over a whole cycle is zero ((d) false).

Teacher's Note:
1. The RMS of a sinusoid is 1/sqrt2 times its peak value.
2. Average velocity over a full period is zero.

Answer: (c) (a), (b), (c) and (d)

Explanation:
1. Resultant amplitude \( A=\sqrt{2^2+4^2+2(2)(4)\cos60^\circ}=\sqrt{28}=2\sqrt7\ \text{m} \).
2. At t = 0: \( x=0+4\sin60^\circ=2\sqrt3\ \text{m} \) ((a) true).
3. \( v_{max}=\omega A=10\times2\sqrt7=20\sqrt7\ \text{m/s} \) ((b) true).
4. \( a_{max}=\omega^2A=100\times2\sqrt7=200\sqrt7\ \text{m/s}^2 \) ((c) true).
5. \( E=\tfrac12m\omega^2A^2=\tfrac12(0.02)(100)(28)=28\ \text{J} \) ((d) true).

Teacher's Note:
1. Add the two SHMs as phasors to get the resultant amplitude.
2. Convert the mass to kilograms (20 g = 0.02 kg) before computing energy.

Answer: (c) first increase and then decrease to the original value

Explanation:
1. As water leaves, the centre of mass of the bob (shell plus water) moves downwards, increasing the effective length and the period.
2. When the water is gone, the centre of mass returns to the centre of the shell, so the period returns to its original value.

Teacher's Note:
1. The effective length is measured to the centre of mass of the bob.
2. The centre of mass first descends and then returns to the geometric centre.

Answer: (b) \( \frac16 \) s

Explanation:
1. KE = 75% of E means PE = 25% of E: \( \frac{x^2}{A^2}=\frac14\Rightarrow x=\frac A2 \).
2. \( x=A\sin\frac{2\pi}{T}t=A\sin\pi t=\frac A2\Rightarrow\pi t=\frac\pi6 \), so \( t=\frac16\ \text{s} \).

Teacher's Note:
1. PE/E = (x/A)^2.
2. Starting from the origin means x = A sin(wt).

Answer: (a) 0.5 s

Explanation:
1. Speed is maximum at the mean position, \( x=0 \).
2. \( \cos\pi t=0\Rightarrow\pi t=\frac\pi2\Rightarrow t=0.5\ \text{s} \).

Teacher's Note:
1. The object is at its extreme position at t = 0 for a cosine displacement.
2. Maximum speed is reached a quarter period later, T/4 with T = 2 s.

Answer: (d) \( \pi^2ma^2\nu^2 \)

Explanation:
1. Average KE over a quarter oscillation equals the average over the whole cycle: \( \overline{KE}=\tfrac14m\omega^2a^2 \).
2. \( =\tfrac14m(2\pi\nu)^2a^2=\pi^2ma^2\nu^2 \).

Teacher's Note:
1. The average of cos^2 or sin^2 over any quarter (or half) cycle is 1/2, so average KE is half of the maximum KE.
2. Maximum KE is (1/2) m w^2 a^2.

Answer: (d) \( \frac\pi3 \)

Explanation:
1. The separation is \( X_0+(x_2-x_1) \), where \( x_2-x_1 \) is an SHM of amplitude \( 2A\sin\frac\phi2 \).
2. Maximum separation \( X_0+2A\sin\frac\phi2=X_0+A \) gives \( \sin\frac\phi2=\frac12 \), so \( \phi=\frac\pi3 \).

Teacher's Note:
1. Subtract the two displacements to get the variable part of the separation.
2. The amplitude of the difference of two equal SHMs is 2A sin(phi/2).

Answer: (c) \( 2\pi\sqrt{\frac{ld}{\rho g}} \)

Explanation:
1. Mass of cube \( m=dl^3 \). If it is pushed down by x, the extra buoyant force is \( \rho\,l^2x\,g \), so the restoring force constant is \( k=\rho gl^2 \).
2. \( T=2\pi\sqrt{\frac mk}=2\pi\sqrt{\frac{dl^3}{\rho gl^2}}=2\pi\sqrt{\frac{ld}{\rho g}} \).

Teacher's Note:
1. The restoring force comes from the extra displaced liquid, equal to rho g A x.
2. Only the cross-sectional area l^2 matters for the force constant.

Answer: (b) 0.729

Explanation:
1. For light damping the amplitude falls exponentially: \( A=A_0e^{-\gamma t} \), so in each equal interval it is multiplied by the same factor.
2. In 5 s the factor is 0.9. The total time after another 10 s is 15 s = three intervals of 5 s.
3. \( \alpha=0.9^3=0.729 \).

Teacher's Note:
1. Exponential decay means equal time intervals give equal ratios.
2. Count the total time from the start (5 + 10 = 15 s), not just the additional 10 s.

Answer: (a) \( \left[\left(\frac{T_M}{T}\right)^2-1\right]\frac{A}{Mg} \)

Explanation:
1. The extra mass stretches the wire: \( \Delta l=\frac{Mgl}{AY} \).
2. Period depends on length: \( \frac{T_M^2}{T^2}=\frac{l+\Delta l}{l}=1+\frac{Mg}{AY} \).
3. \( \frac1Y=\left[\left(\frac{T_M}{T}\right)^2-1\right]\frac{A}{Mg} \).

Teacher's Note:
1. Hooke's law for the wire: Y = (F/A)/(Delta l / l).
2. T squared is proportional to the length of the pendulum.

Answer: (a) \( \frac{7A}{3} \)

Explanation:
1. At \( x=\frac{2A}{3} \): \( v^2=\omega^2\left(A^2-\frac{4A^2}{9}\right)=\frac{5A^2\omega^2}{9} \).
2. New speed \( v'=3v\Rightarrow v'^2=5A^2\omega^2 \) (the angular frequency is unchanged).
3. New amplitude: \( A'^2=x^2+\frac{v'^2}{\omega^2}=\frac{4A^2}{9}+5A^2=\frac{49A^2}{9} \), so \( A'=\frac{7A}{3} \).

Teacher's Note:
1. Angular frequency does not change when the speed changes at fixed position, because it is set by k and m.
2. Use A^2 = x^2 + v^2/w^2 for the new motion.

Answer: (d) \( \frac1{2\pi}\sqrt{\frac{6k}{M}} \)

Explanation:
1. For a small rotation \( \theta \) each end moves \( \frac L2\theta \), so each spring exerts force \( \frac{kL\theta}{2} \) at lever arm \( \frac L2 \).
2. Total restoring torque \( =2\cdot\frac{kL\theta}{2}\cdot\frac L2=\frac{kL^2\theta}{2} \).
3. With \( I=\frac{ML^2}{12} \): \( \omega^2=\frac{kL^2/2}{ML^2/12}=\frac{6k}{M} \), so \( f=\frac1{2\pi}\sqrt{\frac{6k}{M}} \).

Teacher's Note:
1. Use torque about the pivot and the moment of inertia of a rod about its centre.
2. Torsional-type equation I theta'' = -(kL^2/2) theta.

Answer: (c) (a) and (d) only

Explanation:
1. The centre of mass of the system is at the same place in both cases, so the restoring torque is the same: \( \left(\tfrac{mL}{2}+ML\right)g\,\theta \) ((a) true, (b) false).
2. In case B the disc does not rotate with the rod (it can spin freely about its centre), so its moment of inertia contributes only \( ML^2 \). In case A it also contributes \( \tfrac12MR^2 \) about its own centre, so \( I_A\gt I_B \).
3. \( \omega=\sqrt{\tau/I} \), so \( \omega_A\lt \omega_B \) ((d) true, (c) false).

Teacher's Note:
1. Torque depends only on the centre of mass position and total weights; frequency also depends on the moment of inertia.
2. A disc that can spin freely about its centre keeps its orientation while the rod swings.

Answer: (b) \( V_0\gt E\gt 0 \)

Explanation:
1. Potential energy is never negative, so \( E\gt 0 \) is needed for any motion.
2. If \( E\gt V_0 \) the particle reaches the flat region where the force is zero and escapes to infinity, so it is not periodic.
3. Periodic (bound) motion requires \( 0\lt E\lt V_0 \).

Teacher's Note:
1. A particle is bound only if it cannot reach the region where the potential is flat.
2. E cannot be below the minimum potential energy (here zero).

Answer: (b) \( \frac1A\sqrt{\frac m\alpha} \)

Explanation:
1. Dimensional analysis: with \( V=\alpha x^4 \), the only combination of m, \( \alpha \) and A with the dimension of time is \( \sqrt{\frac{m}{\alpha}}\cdot\frac{1}{A} \) (since \( \alpha A^2/m \) has the dimension of \( \text{s}^{-2} \)).
2. Check by energy: \( \tfrac12mv^2=\alpha(A^4-x^4) \) gives \( v\sim A^2\sqrt{\alpha/m} \), so \( T\sim\frac{A}{v}=\frac1A\sqrt{\frac m\alpha} \).

Teacher's Note:
1. Estimate T as amplitude divided by typical speed.
2. A larger amplitude gives a larger speed in a quartic well, so the period decreases with amplitude.

Answer: (a) zero

Explanation:
1. For \( |x|\gt X_0 \) the potential is constant, so the force \( F=-\frac{dV}{dx}=0 \).
2. Hence the acceleration is zero.

Teacher's Note:
1. Force is the negative gradient of potential energy.
2. A flat potential means no force.

Answer: (d) \( A,\ \frac{4\pi}{3} \)

Explanation:
1. The sum of the first two has amplitude \( 2A\cos\frac{\pi}{3}=A \) and phase \( \frac\pi3 \): \( x_1+x_2=A\sin\left(\omega t+\frac\pi3\right) \).
2. To cancel it, \( x_3=-A\sin\left(\omega t+\frac\pi3\right)=A\sin\left(\omega t+\frac\pi3+\pi\right) \).
3. So \( B=A \) and \( \phi=\frac{4\pi}{3} \).

Teacher's Note:
1. Add the first two as phasors; the third must be equal in magnitude and opposite in direction.
2. Opposite means a phase shift of pi.

Answer: (c) \( \sqrt{50} \) m/s

Explanation:
1. Position of the block at t = 1 s: \( x=4.9+0.2\cos\frac\pi3=4.9+0.1=5.0\ \text{m} \) from O.
2. The pebble must cover a horizontal distance \( 10-5=5\ \text{m} \) in 1 s: \( v\cos45^\circ\times1=5\Rightarrow v=5\sqrt2=\sqrt{50}\ \text{m/s} \).
3. Check height: \( v\sin45^\circ(1)-\tfrac12(10)(1)^2=5-5=0 \), so it is back at the level of the block.

Teacher's Note:
1. Find where the block is at t = 1 s using its SHM equation about the natural length.
2. Horizontal and vertical motions of the projectile are independent.

Answer: (c) \( E_1=4E_2 \)

Explanation:
1. The intercept on the position axis is the amplitude: \( A_1=2a \) and \( A_2=a \).
2. For the same oscillator \( E=\tfrac12m\omega^2A^2\propto A^2 \), so \( \frac{E_1}{E_2}=\left(\frac{2a}{a}\right)^2=4 \).

Teacher's Note:
1. The point where the curve meets the position axis has zero momentum, so it is the extreme position (amplitude).
2. Energy is proportional to the square of the amplitude.