Two simple harmonic motions of angular freque | vedclass

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(D) The angular frequencies are given as $\omega_{1} = 100 \, rad \, s^{-1}$ and $\omega_{2} = 1000 \, rad \, s^{-1}$.Let the common displacement ampl

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$1:10^2$

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(D) The angular frequencies are given as $\omega_{1} = 100 \, rad \, s^{-1}$ and $\omega_{2} = 1000 \, rad \, s^{-1}$.Let the common displacement amplitude be $A$.The maximum acceleration $a_{max}$ for a simple harmonic motion is given by the formula $a_{max} = \omega^2 A$.For the first motion,$a_{max1} = \omega_{1}^2 A = (100)^2 A$.For the second motion,$a_{max2} = \omega_{2}^2 A = (1000)^2 A$.The ratio of their maximum accelerations is $\frac{a_{max1}}{a_{max2}} = \frac{\omega_{1}^2 A}{\omega_{2}^2 A} = \frac{\omega_{1}^2}{\omega_{2}^2}$.Substituting the values,we get $\frac{a_{max1}}{a_{max2}} = \frac{(100)^2}{(1000)^2} = \frac{10000}{1000000} = \frac{1}{100}$.Thus,the ratio is $1:10^2$.

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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