A particle of mass 5 ,g is executing S.H.M. w | vedclass

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Question

(A) Given: Mass $m = 5 \,g = 5 	imes 10^{-3} \,kg$,Amplitude $A = 0.3 \,m$,Time period $T = \frac{\pi}{5} \,s$. Angular frequency $\omega = \frac{2\p

Vedclass · Accepted Answer

$0.15$

Vedclass · Answer

(A) Given: Mass $m = 5 \,g = 5 	imes 10^{-3} \,kg$,Amplitude $A = 0.3 \,m$,Time period $T = \frac{\pi}{5} \,s$. Angular frequency $\omega = \frac{2\pi}{T} = \frac{2\pi}{\pi/5} = 10 \,rad/s$. The maximum force acting on a particle in $S.H.M.$ is given by $F_{max} = m\omega^2A$. Substituting the values: $F_{max} = (5 	imes 10^{-3} \,kg) 	imes (10 \,rad/s)^2 	imes (0.3 \,m)$. $F_{max} = 5 	imes 10^{-3} 	imes 100 	imes 0.3 = 5 	imes 10^{-1} 	imes 0.3 = 0.5 	imes 0.3 = 0.15 \,N$.

$A$ particle of mass $5 \,g$ is executing $S.H.M.$ with an amplitude $0.3 \,m$ and period $\frac{\pi}{5} \,s$. The maximum value of the force acting on the particle is (in $\,N$)

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