First slide

Simple hormonic motion

Question

The equation for the displacement of a particle executing SHM is $\large y\, = \,\left\{ {5\sin 2\pi t} \right\}$ cm. Find (i) the velocity at 3cm from the mean position, (ii) acceleration at 0.5s after leaving the mean position.

Easy

A

$\large 8\pi cm{s^{ - 1}}$ , zero
B

$\large 6\pi cm{s^{ - 1}}$ , zero
C

$\large 4\pi cm{s^{ - 1}}$ , zero
D

$\large 2\pi cm{s^{ - 1}}$ , zero

Solution

$\large y\, = \,5\sin 2\pi t$
$\large A\, = \,5cm\,\,\,\,\omega \, = \,2\pi rad/s\,\,\,\,y\, = \,3cm$
$\large v\, = \,\omega \sqrt {{A^2} - {y^2}} \, = \,2\pi \sqrt {25 - 9} \,cm/s$
$\large a\, = \,{\omega ^2}y\, \Rightarrow \,after\,\,0.5\sec$ it is at mean position as
$\large T\, = \,\frac{{2\pi }}{\omega }\sec \, = \,1\sec$
$\large \Rightarrow y\, = \,0\,\,\,\,\,\,\,\,\,\,\,\,\,a\, = \,0$

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