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BITSAT 2013

Exam: BITSAT
Year: 2013
Subject: Physics
Chapter: Oscillations

BITSAT 2013 · Physics · Oscillations

An instantaneous displacement of a simple harmonic oscillator is...

An instantaneous displacement of a simple harmonic oscillator is

\mathrm{x}=\mathrm{A} \cos (\omega \mathrm{t}+\pi / 4).

Its speed will be maximum at time

Options

A $\pi / 4 \omega$
B $\pi / 2 \omega$
C $\pi / \omega$
D $2 \pi \omega$

Correct answer

A. $\pi / 4 \omega$

Step-by-step solution

Velocity,

\mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}=-\mathrm{A} \omega \sin (\omega \mathrm{t}+\pi / 4)

Velocity will be maximum, when

\omega t+\pi / 4=\pi / 2

or

\omega t=\pi / 2-\pi / 4=\pi / 4

or

t=\pi / 4 \omega

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