Two simple harmonic motions of angular freque | vedclass

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

Vedclass · Accepted Answer

$1: 10^2$

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(B) The maximum acceleration $a_{max}$ of a particle executing 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 $A_1$ and $A_2$ are equal $(A_1 = A_2 = A)$,the ratio of the maximum accelerations is:$\frac{a_{max,1}}{a_{max,2}} = \frac{\omega_1^2 A}{\omega_2^2 A} = \left( \frac{\omega_1}{\omega_2} \right)^2$Substituting the given values $\omega_1 = 300 \ rad/s$ and $\omega_2 = 3000 \ rad/s$:$\frac{a_{max,1}}{a_{max,2}} = \left( \frac{300}{3000} \right)^2 = \left( \frac{1}{10} \right)^2 = \frac{1}{100} = 1: 10^2$Therefore,the correct option is $B$.

Two simple harmonic motions of angular frequency $300 \ rad/s$ and $3000 \ rad/s$ have the same amplitude. The ratio of their maximum accelerations is

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