First slide

Simple hormonic motion

Question

The velocity of a particle under going SHM at the mean position is 2ms^–1. Find the velocity of the particle at the point where the displacement from the mean position is equal to half the amplitude.

Easy

A

1.234ms^–1
B

1.732ms^–1
C

2.732ms^–1
D

1.414ms^–1

Solution

$\large At\,\,y\, = \,\frac{A}{2}\, \Rightarrow \,V\, = \,?\,\omega A\, = \,2m/\sec$
$\large V\, = \,\omega \sqrt {{A^2} - {y^2}} \, = \,\omega \sqrt {{A^2} - \frac{{{A^2}}}{4}}$
$\large V\, = \,\frac{{\sqrt 3 A\omega }}{2}\, = \,\frac{{\sqrt 3 }}{2} \times 2\, = \,\sqrt 3 m/\sec$
V = 1.732m/sec

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