JEE Advanced 2021 Paper 1 · Physics · Simple Harmonic Motion · medium

A block of mass 0.50 kg attached to a spring of constant 50 N m^-1 oscillates on a horizontal surface in a medium that exerts a retarding force -bv…

One or more options correct

A block of mass

0.50\ \mathrm{kg}

attached to a spring of constant

50\ \mathrm{N\,m^{-1}}

oscillates on a horizontal surface in a medium that exerts a retarding force

-bv

with

b=1.0\ \mathrm{kg\,s^{-1}}

. It is released from rest at maximum extension. Choose the correct statement(s).

A. The amplitude falls to half of its initial value after about $0.69\ \mathrm{s}$ .
B. The mechanical energy decays with a time constant of $1.0\ \mathrm{s}$ .
C. The angular frequency of the damped oscillation is $9.0\ \mathrm{rad\,s^{-1}}$ .
D. Less than half of the initial mechanical energy is lost during the first complete oscillation.

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Answer: A

The equation

m\ddot x+b\dot x+kx=0

has damping constant

\gamma=b/2m=1.0\ \mathrm{s^{-1}}

and natural frequency

\omega_0=\sqrt{k/m}=10\ \mathrm{rad\,s^{-1}}

. (A) The amplitude is

A_0e^{-\gamma t}

, so it halves at

t=\ln2/\gamma=0.693

s: correct. (B) Energy is proportional to

A^2\propto e^{-2\gamma t}

, so the time constant is

1/(2\gamma)=0.50

s, not

1.0

s: incorrect. (C)

\omega'=\sqrt{\omega_0^2-\gamma^2}=\sqrt{99}=9.95\ \mathrm{rad\,s^{-1}}

, not

9.0

: incorrect. (D) The period is

T'=2\pi/\omega'=0.631

s, and the fraction of energy remaining is

e^{-2\gamma T'}=e^{-1.263}=0.283

, so about

71.7\%

is lost, which is more than half: incorrect. Only (A) is correct.

Original question written in the JEE Advanced 2021 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake