First slide

Simple hormonic motion

Question

A body of mass 36gm moves with S.H.M of amplitude A = 13 cm and period T = 12sec. At a time t = 0 the displacement is x = +13cm. The shortest time of passage from x = +6.5cm to x = – 6.5cm is

Easy

A

4sec
B

2 sec
C

6 sec
D

3 sec

Solution

$\large \omega \, = \,\frac{{2\pi }}{T}$
Equation of motion is $\large y\, = \,A\cos \omega t$
At $\large y\, = \, + \frac{\pi }{2},$ $\large A\cos \omega {t_1}\, = \,\frac{\pi }{2}$
$\large \Rightarrow \,\,\omega {t_1}\, = \,\frac{\pi }{3}$
At $\large y\, = \, - \frac{\pi }{2},A\cos \omega {t_2}\, = \, - \frac{A}{2}$
$\large \Rightarrow \,\omega {t_2}\, = \,{120^o}\, = \,\frac{{2\pi }}{3}$
$\large \therefore \,{t_2} - {t_1}\, = \,\frac{1}{\omega }\left( {\frac{{2\pi }}{3} - \frac{\pi }{3}} \right)\, = \,\frac{T}{{2\pi }} \times \frac{\pi }{3}$
$\large \, = \,\frac{{12}}{{2\pi }} \times \frac{\pi }{3}\,\sec \, = \,2\sec$

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