A particle of mass 1 kg is moving in SHM wit | vedclass

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(D) The path length of $SHM$ is equal to $2A$,where $A$ is the amplitude.Given path length $= 0.01 \ m$,so $2A = 0.01 \ m$,which implies $A = 0.005 \

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$50$

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(D) The path length of $SHM$ is equal to $2A$,where $A$ is the amplitude.Given path length $= 0.01 \ m$,so $2A = 0.01 \ m$,which implies $A = 0.005 \ m$.The frequency $f = 50 \ Hz$.The maximum force $F_{max}$ acting on a particle in $SHM$ is given by $F_{max} = m \omega^2 A$.Since $\omega = 2 \pi f$,we have $F_{max} = m (2 \pi f)^2 A = m (4 \pi^2 f^2) A$.Substituting the values: $F_{max} = 1 	imes 4 	imes \pi^2 	imes (50)^2 	imes 0.005$.$F_{max} = 4 	imes \pi^2 	imes 2500 	imes 0.005$.$F_{max} = 10000 	imes \pi^2 	imes 0.005 = 50 \pi^2 \ N$.

$A$ particle of mass $1 \ kg$ is moving in $SHM$ with a path length of $0.01 \ m$ and frequency of $50 \ Hz$. The maximum force in newton,acting on the particle is (in $\pi^2$)

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