JEE Physics Oscillations MCQs Set 12

Answer: (a) \( 2\pi\sqrt{\frac{M}{2k}} \)

Explanation:
1. When the body is displaced by x along the plane, one spring stretches and the other compresses, both pulling it back.
2. Restoring force \( =-2kx \) (the component of weight only shifts the equilibrium).
3. \( T=2\pi\sqrt{\frac{M}{2k}} \).

Teacher's Note:
1. The weight component along the incline only changes the equilibrium position, not the period.
2. Two springs acting on the same side of displacement add their force constants.

Answer: (b) \( 2\pi\sqrt{\frac{3R}{2g}} \)

Explanation:
1. Moment of inertia about O (parallel axis theorem): \( I=\tfrac12MR^2+MR^2=\tfrac32MR^2 \).
2. For a physical pendulum with the centre of mass at distance R: \( T=2\pi\sqrt{\frac{I}{MgR}}=2\pi\sqrt{\frac{3R}{2g}} \).

Teacher's Note:
1. Physical pendulum: T = 2π√(I/(Mgd)).
2. d is the distance from the pivot to the centre of mass.

Answer: (b) \( 2\pi \)

Explanation:
1. In SHM \( a=-\omega^2x \), so the magnitude of the slope is \( \omega^2=\tan45^\circ=1 \).
2. So \( \omega=1\ \text{rad/s} \) and \( T=\frac{2\pi}{\omega}=2\pi\ \text{s} \).

Teacher's Note:
1. The slope of the a-x graph is −ω².
2. tan 45° = 1.

Answer: (a) 0.6 m

Explanation:
1. \( F=-kx\Rightarrow k=10\ \text{N/m} \), \( \omega=\sqrt{\frac km}=\sqrt{\frac{10}{0.1}}=10\ \text{rad/s} \).
2. Maximum speed \( v=A\omega\Rightarrow A=\frac{6}{10}=0.6\ \text{m} \).

Teacher's Note:
1. The speed is greatest at the mean position and equals Aω.
2. ω = √(k/m).

Answer: (d) \( \frac{\pi}{3\omega} \)

Explanation:
1. At \( x=-\frac A2 \): \( \sin\phi=-\frac12\Rightarrow\phi=-\frac\pi6 \). At \( x=+\frac A2 \): \( \phi=\frac\pi6 \).
2. Phase change \( \frac\pi6-\left(-\frac\pi6\right)=\frac\pi3 \).
3. \( t=\frac{\Delta\phi}{\omega}=\frac{\pi}{3\omega} \).

Teacher's Note:
1. Use the reference circle: time = phase difference / ω.
2. The constant π/6 in the equation does not affect the time between two positions.

Answer: (c) \( \frac T2 \)

Explanation:
1. Each half has twice the force constant: \( 2k \).
2. Two halves in parallel: \( k_{eff}=2k+2k=4k \).
3. \( T'=2\pi\sqrt{\frac{m}{4k}}=\frac T2 \).

Teacher's Note:
1. A shorter spring is stiffer: k is inversely proportional to the length.
2. Parallel springs add their force constants.

Answer: (b) \( \frac\pi{100} \) s

Explanation:
1. Maximum kinetic energy (at the mean position) \( =9-5=4\ \text{J}=\tfrac12m\omega^2A^2 \).
2. \( 4=\tfrac12(2)\omega^2(0.01)^2\Rightarrow\omega=200\ \text{rad/s} \).
3. \( T=\frac{2\pi}{\omega}=\frac{\pi}{100}\ \text{s} \).

Teacher's Note:
1. The potential energy at the mean position is just a reference value; use the kinetic energy there.
2. Total energy = ½mω²A² measured above the minimum potential energy.

Answer: (d) \( \frac1{T^2}=\frac1{T_1^2}+\frac1{T_2^2} \)

Explanation:
1. Parallel combination: \( k=k_1+k_2 \).
2. Since \( k=\frac{4\pi^2m}{T^2} \): \( \frac{4\pi^2m}{T^2}=\frac{4\pi^2m}{T_1^2}+\frac{4\pi^2m}{T_2^2} \).
3. \( \frac1{T^2}=\frac1{T_1^2}+\frac1{T_2^2} \).

Teacher's Note:
1. Write the force constant in terms of the period.
2. Spring constants add in parallel.

Answer: (a) \( \frac{3T}{4} \)

Explanation:
1. For half of the oscillation the length is ℓ, taking time \( \frac T2 \).
2. For the other half the string pivots about P, so the effective length is \( \ell-\frac{3\ell}4=\frac\ell4 \), giving a period \( \frac T2 \) and a half-period \( \frac T4 \).
3. Total time \( =\frac T2+\frac T4=\frac{3T}{4} \).

Teacher's Note:
1. The period depends on the square root of the length.
2. Each half swing takes half the period of the corresponding length.

Answer: (d) simple harmonic with time period 0.1 s

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

Teacher's Note:
1. Product of sine and cosine of the same angle is a sine of the double angle.
2. Compare with y = A sin(ωt + φ).

Answer: (d) \( 2\pi\sqrt{\frac{da}{g}} \)

Explanation:
1. For a downward displacement x the extra buoyant force is \( F=\rho_wg\,(bc)\,x \).
2. Mass of the block \( m=d\rho_w\,abc \).
3. \( T=2\pi\sqrt{\frac{m}{\rho_wgbc}}=2\pi\sqrt{\frac{da}{g}} \).

Teacher's Note:
1. Restoring force is the extra buoyancy proportional to the displacement.
2. The effective force constant is ρ g × (area of the base).

Answer: (c) \( 2\pi\left(\frac{\sigma L}{\rho g}\right)^{1/2} \)

Explanation:
1. Let the base area be A. Mass \( m=\sigma AL \).
2. Extra buoyant force for a displacement x: \( \rho gAx \), so \( k=\rho gA \).
3. \( T=2\pi\sqrt{\frac{\sigma AL}{\rho gA}}=2\pi\left(\frac{\sigma L}{\rho g}\right)^{1/2} \).

Teacher's Note:
1. The base area cancels.
2. Mass is density times volume.

Answer: (b) \( mg\ell(1-\cos\theta) \)

Explanation:
1. Maximum kinetic energy occurs at the lowest point and equals the loss of potential energy from the extreme position.
2. The bob falls through \( \ell(1-\cos\theta) \).
3. \( K_{max}=mg\ell(1-\cos\theta) \).

Teacher's Note:
1. Energy conservation between the extreme and the lowest position.
2. Height drop = ℓ − ℓ cosθ.

Answer: (a) \( \frac\pi5\sqrt{\frac32} \) s

Explanation:
1. Force constant \( k=\frac{10}{0.05}=200\ \text{N/m} \). Mass \( m=\frac{30}{10}=3\ \text{kg} \).
2. \( T=2\pi\sqrt{\frac mk}=2\pi\sqrt{\frac3{200}}=\frac\pi5\sqrt{\frac32}\ \text{s} \) (about 0.77 s).

Teacher's Note:
1. Find k from force and extension, and m from the weight.
2. T = 2π√(m/k).

Answer: (d) 0.1 s

Explanation:
1. Measured from the extreme position: \( x=a\cos\omega t \). After travelling a/2 the displacement from the mean is \( \frac a2 \).
2. \( \cos\omega t=\frac12\Rightarrow\omega t=\frac\pi3 \).
3. \( t=\frac{T}{6}=\frac{0.60}{6}=0.1\ \text{s} \).

Teacher's Note:
1. Use x = a cos ωt when starting from the extreme position.
2. Travelling a/2 from the extreme leaves a/2 from the mean.

Answer: (c) 0.5 m/s

Explanation:
1. The time for a/2 is \( t=\frac T6=0.1\ \text{s} \) (from the previous result).
2. Mean velocity \( =\frac{\text{displacement}}{\text{time}}=\frac{0.05}{0.1}=0.5\ \text{m/s} \).

Teacher's Note:
1. The point moves in one direction, so displacement equals distance.
2. Mean velocity is total displacement divided by total time.

Answer: (d) A is false, but R is true.

Explanation:
1. From the mean position: \( A\sin\omega t_1=\frac{A}{\sqrt2}\Rightarrow\omega t_1=\frac\pi4 \).
2. From A/√2 to A: the total phase is π/2, so \( \omega t_2=\frac\pi2-\frac\pi4=\frac\pi4 \).
3. Hence t1 = t2, so A is false. R is a correct statement.

Teacher's Note:
1. Use the reference circle to compare phases.
2. The two quarter-phase intervals are equal.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. U = ½kx² and K = ½k(A² − x²) are both quadratic in x, so each graph is a parabola; A is true.
2. R is also true, but it is a consequence of the same fact rather than an explanation of why the shape is a parabola.

Teacher's Note:
1. A parabola is a quadratic curve.
2. Quadratic dependence also means not linear.

Answer: (c) A is true, but R is false.

Explanation:
1. In SHM the acceleration \( a=-\omega^2x \) changes with position, so the motion is not uniformly accelerated; A is true.
2. R is false: for example, in projectile motion the velocity is non-uniform but the acceleration is uniform.

Teacher's Note:
1. Uniform acceleration means constant acceleration, not constant velocity.
2. Projectile motion is a counter-example for R.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. For SHM \( T=2\pi\sqrt{m/k} \), which does not contain the amplitude.
2. So amplitude and period are independent, and R explains A.

Teacher's Note:
1. The period depends on m and k only.
2. Isochronism of SHM.