Question

section a - multiple choice (40 marks)

1. a simple pendulum is oscillating in a lift. if the lift starts moving upwards with a uniform acceleration, then the period will

a. remain the same
b. be shorter
c. be longer
d. may be shorter or longer, depending on the magnitude of acceleration

1. the bob of a simple pendulum of period t is given a negative charge. if it is allowed to oscillate above a positively charged plate, the new time period will be

a. equal to t
b. more than t
c. less than t
d. infinite

1. a body of mass 5 g is executing s.h.m. with amplitude 10 cm. its maximum velocity is 100 cm/s. its velocity will be 50 cm/s at a displacement from the mean position equal to

a. 5 cm
b. 5√3 cm
c. 10 cm
d. 10√3 cm

1. a simple harmonic oscillator has a period of 0.01 s, and an amplitude of 0.2 m. the magnitude of the velocity in m/s at the center of oscillation is

a. 20π
b. 40π
c. 60π
d. 80π

1. a loaded spring vibrates with a period t. the spring is divided into four equal parts and the same load is suspended from one of these parts. the new time period is

a. t/4
b. t/2
c. 2t
d. 4t

1. the length of a seconds pendulum on the surface of the moon, where g is 1/6th of the value of the surface of the earth, is

a. 1/36 m
b. 1/6 m
c. 6 m
d. 36 m

1. a particle of mass 0.5 kg executes s.h.m. its energy is 0.04 j. if its time period is π sec amplitude is

a. 10 cm
b. 15 cm
c. 20 cm
d. 40 cm

Explanation:

Step1: Recall the formula for the period of a simple pendulum

The formula for the period of a simple pendulum is \(T = 2\pi\sqrt{\frac{l}{g_{eff}}}\). When the lift moves upwards with uniform acceleration \(a\), the effective acceleration \(g_{eff}=g + a\). Since \(g_{eff}\) increases, and \(T\propto\frac{1}{\sqrt{g_{eff}}}\), the period \(T\) will be shorter.

Step2: Analyze the effect of charge on the period of a simple pendulum

The period of a simple pendulum \(T = 2\pi\sqrt{\frac{l}{g}}\) is independent of the mass and charge of the bob. So, when the bob is given a charge and oscillates above a charged plate, the new time period will be equal to \(T\).

Step3: Use the formula for velocity in SHM

The formula for velocity in SHM is \(v=\omega\sqrt{A^{2}-x^{2}}\), and also \(v_{max}=\omega A\). Given \(v_{max} = 100\space cm/s\), \(A = 10\space cm\), so \(\omega=\frac{v_{max}}{A}=10\space s^{-1}\). Given \(v = 50\space cm/s\), then \(50=10\sqrt{10^{2}-x^{2}}\). Squaring both sides: \(2500 = 100(100 - x^{2})\), \(25=100 - x^{2}\), \(x^{2}=75\), \(x = 5\sqrt{3}\space cm\).

Step4: Use the formula for maximum velocity in SHM

The formula for maximum velocity in SHM is \(v_{max}=\omega A\), and \(\omega=\frac{2\pi}{T}\). Given \(T = 0.01\space s\), \(A=0.2\space m\), then \(\omega=\frac{2\pi}{0.01}=200\pi\space rad/s\), \(v_{max}=200\pi\times0.2 = 40\pi\space m/s\).

Step5: Recall the formula for the period of a spring - mass system

The formula for the period of a spring - mass system is \(T = 2\pi\sqrt{\frac{m}{k}}\). When a spring is divided into \(n\) equal parts, the spring constant of each part \(k_{new}=nk\). Here \(n = 4\), so \(k_{new}=4k\). The new period \(T_{new}=2\pi\sqrt{\frac{m}{4k}}=\frac{T}{2}\).

Step6: Use the formula for the period of a second's pendulum

A second's pendulum has \(T = 2s\) on earth. The formula \(T = 2\pi\sqrt{\frac{l}{g}}\). On earth \(T_{e}=2\pi\sqrt{\frac{l_{e}}{g_{e}}}=2s\). On the moon \(T_{m}=2\pi\sqrt{\frac{l_{m}}{g_{m}}}\), and \(g_{m}=\frac{g_{e}}{6}\). If we assume the period on the moon is also considered for a pendulum with the same time - keeping concept (assuming \(T_{m}\) is related to the same kind of'second' - like measurement in terms of formula), and if we consider the ratio \(\frac{T_{e}}{T_{m}}=\sqrt{\frac{l_{e}g_{m}}{l_{m}g_{e}}}\). But if we consider the formula \(l=\frac{gT^{2}}{4\pi^{2}}\), for a second's pendulum \(T = 2s\). On earth \(l_{e}=\frac{g_{e}\times4}{4\pi^{2}}\), on the moon \(l_{m}=\frac{g_{m}\times4}{4\pi^{2}}\), and since \(g_{m}=\frac{g_{e}}{6}\), \(l_{m}=\frac{l_{e}}{6}\).

Step7: Use the formula for the energy of a SHM system

The energy of a SHM system is \(E=\frac{1}{2}m\omega^{2}A^{2}\), and \(\omega=\frac{2\pi}{T}\). Given \(m = 0.5\space kg\), \(E = 0.04\space J\), \(T=\pi\space s\), \(\omega=\frac{2\pi}{\pi}=2\space rad/s\). Then \(0.04=\frac{1}{2}\times0.5\times4\times A^{2}\), \(0.04 = A^{2}\), \(A = 0.2\space m=20\space cm\).

Answer:

1. B. be shorter
2. A. equal to \(T\)
3. B. \(5\sqrt{3}\space cm\)
4. B. \(40\pi\)
5. B. \(T/2\)
6. B. \(1/6\space m\)
7. C. \(20\space cm\)