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

Vedclass · Accepted Answer

$0.15$

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(B) The maximum velocity $(v_{\max})$ of a particle executing simple harmonic motion is given by the formula $v_{\max} = A \omega$, where $A$ is the amplitude and $\omega$ is the angular frequency.Given: Amplitude $A = 50 \,mm = 50 	imes 10^{-3} \,m = 0.05 \,m$.Time period $T = 2 \,s$.The angular frequency is $\omega = \frac{2 \pi}{T} = \frac{2 \pi}{2} = \pi \,rad/s$.Substituting the values: $v_{\max} = 0.05 	imes \pi \approx 0.05 	imes 3.14159 = 0.157 \,ms^{-1}$.Rounding to two decimal places, we get $v_{\max} \approx 0.16 \,ms^{-1}$. However, based on the provided options, the closest value is $0.15 \,ms^{-1}$.

If a simple pendulum oscillates with an amplitude of $50 \,mm$ and time period of $2 \,s$, then its maximum velocity is (in $\,ms^{-1}$)

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