Q108 / 125 · BITSAT 2013 · Physics

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

  • Correct answer
    A
    $\pi / 4 \omega$
  • B
    $\pi / 2 \omega$
  • C
    $\pi / \omega$
  • D
    $2 \pi \omega$

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 $t=\pi / 4 \omega$

Hence, the correct answer is option: 'A'.

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