A particle of mass 10 ,g is undergoing S.H.M. | vedclass

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(D) Given: Mass $m = 10 \,g = 0.01 \,kg$,Amplitude $A = 10 \,cm = 0.1 \,m$,Period $T = 0.1 \,s$.The angular frequency $\omega$ is given by $\omega = \

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$4$

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(D) Given: Mass $m = 10 \,g = 0.01 \,kg$,Amplitude $A = 10 \,cm = 0.1 \,m$,Period $T = 0.1 \,s$.The angular frequency $\omega$ is given by $\omega = \frac{2\pi}{T} = \frac{2\pi}{0.1} = 20\pi \,rad/s$.The maximum acceleration in $S.H.M.$ is $a_{	ext{max}} = \omega^2 A$.The maximum force is $F_{	ext{max}} = m \cdot a_{	ext{max}} = m \omega^2 A$.Substituting the values: $F_{	ext{max}} = 0.01 	imes (20\pi)^2 	imes 0.1$.$F_{	ext{max}} = 0.001 	imes 400 \pi^2 = 0.4 \pi^2$.Using $\pi^2 \approx 9.87$,$F_{	ext{max}} \approx 0.4 	imes 9.87 = 3.948 \,N$.Rounding to the nearest integer,$F_{	ext{max}} \approx 4 \,N$.

$A$ particle of mass $10 \,g$ is undergoing $S.H.M.$ with an amplitude of $10 \,cm$ and a period of $0.1 \,s$. The maximum value of the force on the particle is about ............ $N$.

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