MCQ
The ratio of maximum acceleration to maximum velocity in a simple harmonic motion is $10\,s^{-1}$.  At, $t = 0$ the displacement is $5\, m$. What is the maximum acceleration ? The initial phase is $\frac{\pi }{4}$
  • A
    $500\, m/s^2$
  • $500\,\sqrt 2 \,m/{s^2}$
  • C
    $750\, m/s^2$
  • D
    $750\,\sqrt 2 \,m/{s^2}$

Answer

Correct option: B.
$500\,\sqrt 2 \,m/{s^2}$
b
Maximum velocity in $\mathrm{SHM}, \mathrm{v}_{\max }=\mathrm{a} \omega$
Maximum acceleration in $\mathrm{SHM}, \mathrm{A}_{\max }=\mathrm{a} \omega^{2}$
where $a$ and $\omega$ are maximum amplitude and angular frequency.
Given that, $\frac{A_{\max }}{v_{\max }}=10$
i.e., $\omega=10 \mathrm{s}^{-1}$
Displacement is given by

$x=a \sin (\omega t+\pi t 4)$
at $t=0, x=5$

$5=a \sin \pi / 4$

$5=a \sin 45^{\circ} \Rightarrow a=5 \sqrt{2}$
Maximum acceleration $\mathrm{A}_{\mathrm{max}}=\mathrm{a} \omega^{2}$

$=500\sqrt{2}\,m/s^2$

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