JEE Physics Oscillations MCQs Set 15

Answer: (b) \( 5\pi \) m/s

Explanation:
1. Compare with \( x = A\sin(\omega t+\phi) \): \( A = 5 \) m, \( \omega = \pi \) rad/s, so the period is 2 s.
2. \( v_{max} = A\omega = 5\pi \) m/s.

Teacher's Note:
1. Maximum speed is amplitude times angular frequency.
2. The phase constant pi/6 does not affect the maximum speed.

Answer: (a) 100 N/m and \( \sqrt3 \) m/s

Explanation:
1. \( k = m\omega^2 = 0.01\times100^2 = 100 \) N/m.
2. \( v = A\omega\cos(\omega t+\phi) \); at t = 0: \( v = 0.02\times100\times\cos30^\circ = \sqrt3 \) m/s (and \( x(0) = 1.0 \) cm).

Teacher's Note:
1. Convert mass to kilograms and amplitude to metres.
2. Differentiate x to get v, then put t equal to zero.

Answer: (d) T/12

Explanation:
1. Take \( x = A\sin\omega t \). \( x = A/2 \) when \( \sin\omega t = 1/2 \), i.e. \( \omega t = \pi/6 \).
2. \( t = \dfrac{\pi/6}{2\pi/T} = T/12 \).

Teacher's Note:
1. Use the phasor (reference circle) picture: A/2 corresponds to 30 degrees.
2. T/12 corresponds to 30 degrees of the circle.

Answer: (c) T/6

Explanation:
1. Take \( x = A\cos\omega t \). \( x = A/2 \) at \( \omega t = \pi/3 \).
2. \( t = \dfrac{\pi/3}{2\pi/T} = T/6 \).

Teacher's Note:
1. On the reference circle the particle moves through 60 degrees.
2. Use cosine form because the motion starts at the extreme.

Answer: (b) T/4

Explanation:
1. \( x = A\sin\omega t = \pm A/\sqrt2 \) at \( \omega t = \mp\pi/4 \).
2. The angle swept is \( \pi/2 \), so \( t = T/4 \).

Teacher's Note:
1. Each half-way point is 45 degrees from the mean position.
2. The total angle is 90 degrees, a quarter of the circle.

Answer: (a) \( \dfrac{\sqrt3}{2}v_0 \)

Explanation:
1. \( v = \omega\sqrt{A^2-x^2} = \omega\sqrt{A^2-A^2/4} = \dfrac{\sqrt3}{2}\omega A \).
2. Since \( v_0 = \omega A \), \( v = \dfrac{\sqrt3}{2}v_0 \).

Teacher's Note:
1. Use v = omega times the square root of (A squared minus x squared).
2. Energy conservation gives the same result.

Answer: (d) \( \dfrac{10\pi^2}{9} \) cm/s\(^2\) (about 11 cm/s\(^2\))

Explanation:
1. \( \omega = 2\pi/6 = \pi/3 \) rad/s. At t = 0, \( 5 = 10\sin\phi \) and the particle moves towards +x, so \( \phi = \pi/6 \): \( x = 10\sin(\pi t/3+\pi/6) \) cm.
2. At t = 4 s: \( x = 10\sin(3\pi/2) = -10 \) cm (the extreme position).
3. \( |a| = \omega^2|x| = \dfrac{\pi^2}{9}\times10 = \dfrac{10\pi^2}{9} \approx 11 \) cm/s\(^2\).

Teacher's Note:
1. Find the phase constant from the initial position and direction of motion.
2. At the extreme position the acceleration has its maximum magnitude.

Answer: (c) \( \pm1.2 \) cm from the mean position

Explanation:
1. \( \omega = a_{max}/v_{max} = 5 \) rad/s and \( A = v_{max}/\omega = 2 \) cm.
2. \( v^2 = \omega^2(A^2-x^2) \Rightarrow 64 = 25(4-x^2) \Rightarrow x^2 = 1.44 \), so \( x = \pm1.2 \) cm.

Teacher's Note:
1. Two maxima give omega and A directly.
2. Then use the velocity formula.

Answer: (b) 5 cm

Explanation:
1. \( \tfrac12k(A^2-x^2) = 3\times\tfrac12kx^2 \Rightarrow A^2 = 4x^2 \).
2. So x = A/2 = 5 cm.

Teacher's Note:
1. Kinetic energy is (1/2) k (A squared minus x squared).
2. Equal energies occur at A over root 2; three times needs A/2.

Answer: (d) about 0.1 N/m (\( \pi^2/100 \))

Explanation:
1. \( T = 2\pi\sqrt{m/k} \Rightarrow k = \dfrac{4\pi^2m}{T^2} = \dfrac{4\pi^2\times0.01}{4} = \dfrac{\pi^2}{100} \)
2. This is about 0.099 N/m.

Teacher's Note:
1. Convert 10 g to 0.01 kg first.
2. Rearrange the time period formula for k.

Answer: (c) about 40 J

Explanation:
1. \( k = 4\pi^2m/T^2 = \pi^2/2 \) N/m.
2. Extension in equilibrium \( x = mg/k \); energy \( U = \tfrac12kx^2 = \dfrac{(mg)^2}{2k} = \dfrac{400}{\pi^2}\approx 40 \) J.

Teacher's Note:
1. First get k from the time period.
2. Energy stored equals (mg) squared over 2k.

Answer: (b) about 0.16 kg

Explanation:
1. \( \tfrac12kx^2 = 5 \Rightarrow k = \dfrac{10}{0.0625} = 160 \) N/m.
2. \( \omega = 2\pi\times5 = 10\pi \) rad/s, so \( m = k/\omega^2 = 160/(100\pi^2) = 1.6/\pi^2 \approx 0.16 \) kg.

Teacher's Note:
1. Energy stored fixes the spring constant.
2. The frequency then gives the mass.

Answer: (a) \( \dfrac{F^2}{2k} \)

Explanation:
1. The mean (equilibrium) position is where \( kx_0 = F \), so \( x_0 = F/k \) (this is also the amplitude).
2. Energy stored in the spring \( = \tfrac12kx_0^2 = \dfrac{F^2}{2k} \). (Kinetic energy at that point is also \( F^2/2k \).)

Teacher's Note:
1. The oscillation is about the point where the spring force balances F.
2. The amplitude equals the distance from the unstretched position to that point.

Answer: (d) \( 2\pi\sqrt{\dfrac{11m}{6k}} \)

Explanation:
1. \( \dfrac1{k_{eq}} = \dfrac1k+\dfrac1{2k}+\dfrac1{3k} = \dfrac{11}{6k} \), so \( k_{eq} = \dfrac{6k}{11} \).
2. \( T = 2\pi\sqrt{m/k_{eq}} = 2\pi\sqrt{\dfrac{11m}{6k}} \).

Teacher's Note:
1. Springs in series add as reciprocals.
2. Gravity only shifts the mean position; it does not change the period.

Answer: (b) 0.25 m

Explanation:
1. \( \omega = 2\pi \) rad/s, so \( T = 1 \) s.
2. \( L = \dfrac{gT^2}{4\pi^2} = \dfrac{\pi^2}{4\pi^2} = 0.25 \) m.

Teacher's Note:
1. Read omega from the argument of the sine.
2. The amplitude (1 degree) is irrelevant to the period.

Answer: (a) 9.794 m/s\(^2\)

Explanation:
1. A slow clock loses 24 s in 86400 s, so its period is longer by the ratio \( T'/T = \dfrac{86400}{86376} = \dfrac{3600}{3599} \).
2. \( g' = g(T/T')^2 = 9.8\times\left(\dfrac{3599}{3600}\right)^2 \approx 9.794 \) m/s\(^2\).

Teacher's Note:
1. A slow clock means a longer period, which means a smaller g.
2. Compare the number of ticks in a day.

Answer: (d) \( 2T_0 \)

Explanation:
1. Effective gravity \( g_{eff} = g-\tfrac{3g}{4} = \tfrac g4 \).
2. \( T = 2\pi\sqrt{L/g_{eff}} = 2\times2\pi\sqrt{L/g} = 2T_0 \).

Teacher's Note:
1. In a downward-accelerating lift the effective gravity is reduced.
2. T is inversely proportional to the square root of g.

Answer: (b) \( 2\pi\sqrt{\dfrac{7\ell}{12g}} \)

Explanation:
1. \( I = \dfrac{m\ell^2}{12}+m\left(\dfrac\ell4\right)^2 = \dfrac{7m\ell^2}{48} \) and \( d = \ell/4 \).
2. \( T = 2\pi\sqrt{\dfrac{I}{mgd}} = 2\pi\sqrt{\dfrac{7\ell^2/48}{g\ell/4}} = 2\pi\sqrt{\dfrac{7\ell}{12g}} \).

Teacher's Note:
1. Use the parallel axis theorem for the moment of inertia.
2. Period of a compound pendulum is 2 pi times the root of I over (m g d).

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

Explanation:
1. \( I = mr^2+mr^2 = 2mr^2 \) about the point of suspension; \( d = r \).
2. \( T = 2\pi\sqrt{\dfrac{2mr^2}{mgr}} = 2\pi\sqrt{\dfrac{2r}{g}} \).

Teacher's Note:
1. A ring behaves like a simple pendulum of length 2r.
2. Moment of inertia about a tangent is 2 m r squared.

Answer: (d) \( 2\pi\sqrt{\dfrac{2\sqrt2\,a}{3g}} \)

Explanation:
1. About the centre, \( I_c = \dfrac{ma^2}{6} \); the corner is at \( d = \dfrac{a}{\sqrt2} \), so \( I = \dfrac{ma^2}{6}+\dfrac{ma^2}{2} = \dfrac{2ma^2}{3} \).
2. \( T = 2\pi\sqrt{\dfrac{2a^2/3}{g\,a/\sqrt2}} = 2\pi\sqrt{\dfrac{2\sqrt2\,a}{3g}} = 2\pi\sqrt{\dfrac{\sqrt8\,a}{3g}} \).

Teacher's Note:
1. The centre of mass is half the diagonal from a corner.
2. Use the parallel axis theorem with the square's moment of inertia about its centre.

Answer: (c) \( 2\pi\sqrt{\dfrac{3r}{2g}} \)

Explanation:
1. \( I = \dfrac{mr^2}{2}+m\left(\dfrac r2\right)^2 = \dfrac{3mr^2}{4} \) and \( d = r/2 \).
2. \( T = 2\pi\sqrt{\dfrac{3r^2/4}{g\,r/2}} = 2\pi\sqrt{\dfrac{3r}{2g}} \).

Teacher's Note:
1. Moment of inertia of a disc about its centre is m r squared over 2.
2. d is the distance from the point of suspension to the centre.

Answer: (b) \( r/\sqrt2 \) and \( 2\pi\sqrt{\dfrac{\sqrt2\,r}{g}} \)

Explanation:
1. \( T = 2\pi\sqrt{\dfrac{k^2+d^2}{gd}} \) with radius of gyration \( k = r/\sqrt2 \) about the centre. It is minimum when \( d = k = r/\sqrt2 \).
2. \( T_{min} = 2\pi\sqrt{\dfrac{2k}{g}} = 2\pi\sqrt{\dfrac{\sqrt2\,r}{g}} \).

Teacher's Note:
1. The period is smallest when the distance d equals the radius of gyration k.
2. Then (k squared plus d squared) over d equals 2k.

Answer: (a) \( 2\pi\sqrt{\dfrac{L}{\sqrt3\,g}} \) at \( \dfrac{L}{2\sqrt3} \)

Explanation:
1. Radius of gyration about the centre: \( k = L/\sqrt{12} = L/(2\sqrt3) \). The minimum period occurs at \( d = k \).
2. \( T_{min} = 2\pi\sqrt{\dfrac{2k}{g}} = 2\pi\sqrt{\dfrac{L}{\sqrt3\,g}} \).

Teacher's Note:
1. Use k squared equal to L squared over 12 for a thin rod.
2. The minimum period needs d equal to k.

Answer: (d) \( \sqrt{37} \) cm (about 6.1 cm)

Explanation:
1. \( A = \sqrt{a_1^2+a_2^2+2a_1a_2\cos\phi} = \sqrt{9+16+2(3)(4)(0.5)} \)
2. \( = \sqrt{37} \approx 6.1 \) cm.

Teacher's Note:
1. The resultant amplitude is the length of the sum of two phasors.
2. Phi zero gives 7 cm and phi 90 degrees gives 5 cm.

Answer: (a) about 6.93 s

Explanation:
1. Amplitude \( A = A_0e^{-bt/2m} \). Setting \( A = A_0/2 \): \( t = \dfrac{2m\ln2}{b} \).
2. \( t = \dfrac{2\times0.2\times0.693}{0.04} = 6.93 \) s.

Teacher's Note:
1. Energy decays at twice the rate of the amplitude.
2. Convert b to kg/s: 40 g/s is 0.04 kg/s.

Answer: (d) about 3.47 s

Explanation:
1. Energy \( E = E_0e^{-bt/m} \). Setting \( E = E_0/2 \): \( t = \dfrac{m\ln2}{b} \).
2. \( t = \dfrac{0.2\times0.693}{0.04} = 3.47 \) s.

Teacher's Note:
1. Energy depends on amplitude squared, so it decays twice as fast as the amplitude.
2. The period of this oscillator is about 0.31 s, so many oscillations occur in 3.47 s.

Answer: (c) As x increases c also increases

Explanation:
1. SHM requires \( F = -kx \) with constant k, so \( cx^{1/3} = kx \).
2. Therefore \( c = kx^{2/3} \), which increases with x.

Teacher's Note:
1. A force of the form -c x to the one third is not linear unless c itself depends on x.
2. Solve for c as a function of x.

Answer: (b) an extreme position

Explanation:
1. At any point other than an extreme the particle passes twice per period (once in each direction), and the two time gaps are unequal unless the point is the mean position, where the gaps are T/2.
2. At an extreme position the particle appears only once per cycle, so successive appearances are exactly T apart.

Teacher's Note:
1. Count how many times the particle passes a point in one period.
2. The mean position gives equal gaps of T/2.

Answer: (a) \( V_{max} \)

Explanation:
1. The velocity \( V_{max}\cos\omega t \) takes any value other than \( \pm V_{max} \) twice per period at unequal intervals, so the smallest interval between two occurrences is less than T.
2. Only \( V_{max} \) (in a given direction) occurs once per period, and a zero velocity occurs twice per period (interval T/2).

Teacher's Note:
1. The velocity as a vector (with sign) must repeat once per cycle.
2. Only the extreme value of the velocity occurs once per cycle.