A small body of mass 0.10 , kg is executing S | vedclass

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(A) The maximum force in $S.H.M.$ is given by $F_{\max} = m \cdot a_{\max}$,where $a_{\max} = A \omega^2$.Given: mass $m = 0.10 \, kg$,amplitude $A =

Vedclass · Accepted Answer

$98.596$

Vedclass · Answer

(A) The maximum force in $S.H.M.$ is given by $F_{\max} = m \cdot a_{\max}$,where $a_{\max} = A \omega^2$.Given: mass $m = 0.10 \, kg$,amplitude $A = 1.0 \, m$,and period $T = 0.20 \, s$.The angular frequency $\omega$ is given by $\omega = \frac{2\pi}{T} = \frac{2\pi}{0.20} = 10\pi \, rad/s$.The maximum acceleration is $a_{\max} = A \omega^2 = 1.0 	imes (10\pi)^2 = 100\pi^2 \, m/s^2$.The maximum force is $F_{\max} = m \cdot a_{\max} = 0.10 	imes 100\pi^2 = 10\pi^2 \, N$.Using $\pi^2 \approx 9.8596$,we get $F_{\max} = 10 	imes 9.8596 = 98.596 \, N$.

$A$ small body of mass $0.10 \, kg$ is executing $S.H.M.$ of amplitude $1.0 \, m$ and period $0.20 \, s$. The maximum force acting on it is .... $N$.

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