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Q. The displacement of a particle in SHM is $x=10 sin \left(2t-\frac{\pi}{6}\right)$ metre. When its displacement is $6\, m$, the velocity of the particle (in $ms^{-1}$) is

KEAMKEAM 2013Oscillations

Solution:

Given, $ x=10 \,\sin \left(2 t-\frac{\pi}{6}\right)$
$x=6\, m$
Here, $\omega=2$ and $A=10$
We know that for SHM,
$v=\omega \sqrt{A^{2}-x^{2}} $
$v=2 \sqrt{10^{2}-6^{2}} $
$v=2 \times 8 $
$v=16 \,ms ^{-1}$