Two simple harmonic motions of angular freque | vedclass

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(B) The maximum acceleration of a particle in simple harmonic motion is given by the formula $a_{\max} = \omega^{2} A$,where $\omega$ is the angular f

Vedclass · Accepted Answer

$1: 10^{2}$

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(B) The maximum acceleration of a particle in simple harmonic motion is given by the formula $a_{\max} = \omega^{2} A$,where $\omega$ is the angular frequency and $A$ is the amplitude.Given that the amplitudes $A$ are the same for both motions,the ratio of their maximum accelerations is:$\frac{(a_{\max})_{1}}{(a_{\max})_{2}} = \frac{\omega_{1}^{2} A}{\omega_{2}^{2} A} = \frac{\omega_{1}^{2}}{\omega_{2}^{2}}$Substituting the given values $\omega_{1} = 100 \ rad \ s^{-1}$ and $\omega_{2} = 1000 \ rad \ s^{-1}$:$\frac{(a_{\max})_{1}}{(a_{\max})_{2}} = \frac{(100)^{2}}{(1000)^{2}} = \left(\frac{100}{1000}\right)^{2} = \left(\frac{1}{10}\right)^{2} = \frac{1}{100} = 1: 10^{2}$

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

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