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Q. Two simple harmonic motions of angular frequency $ 100 $ and $ 1000\, rad\, s^{-1} $ have the same displacement amplitude. The ratio of their maximum acceleration is

MHT CETMHT CET 2008

Solution:

Acceleration of simple harmonic motion is
$a_{\max }=-\omega^{2} A$
or $\frac{\left(a_{\max }\right)_{1}}{\left(a_{\max }\right)_{2}}=\frac{\omega_{1}^{2}}{\omega_{2}^{2}}$
(as $A$ remains same)
or $\frac{\left(a_{\max }\right)_{1}}{\left(a_{\max }\right)_{2}}=\frac{(100)^{2}}{(1000)^{2}}=\left(\frac{1}{10}\right)^{2}$
$=1: 10^{2}$