Q. A particle executes simple harmonic motion with an amplitude of 5 cm. When the particle is at 4 cm from the mean position, the magnitude of its velocity in SI units is equal to that of its acceleration. Then, its periodic time in seconds is :

Solution:

$v =\omega\sqrt{A^{2} -x^{2}}$ ....(1)
$ a -\omega^{2} x $ ....(2)
$ \left|v\right| =\left|a\right|$ .....(3)
$ \omega\sqrt{A^{2} -x^{2}} = \omega^{2}x $
$ A^{2} -x^{2} = \omega^{2} x^{2} $
$ 5^{2} -4^{2} =\omega^{2}\left(4^{2}\right) $
$ \Rightarrow 3 = \omega\times4 $
$ T = 2\pi/\omega$

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