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(B) The maximum velocity $(v_{\max})$ of a simple harmonic oscillator is given by the formula: $v_{\max} = A\omega$,where $A$ is the amplitude and $\o

Vedclass · Accepted Answer

$0.16$

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(B) The maximum velocity $(v_{\max})$ of a simple harmonic oscillator 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: $v_{\max} = 0.05 	imes \pi$.Using $\pi \approx 3.14159$,we get $v_{\max} = 0.05 	imes 3.14159 \approx 0.157\, m/s$.Rounding to two decimal places,we get $0.16\, m/s$.Thus,the correct option is $B$.

If a simple pendulum oscillates with an amplitude of $50\, mm$ and time period of $2\, s$,then its maximum velocity is .... $m/s$.

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