A spring executes S.H.M. with a mass of 1 ,kg | vedclass

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(B) Given: Mass $m = 1 \,kg$, Force constant $k = 4 \,N/m$, Velocity $v = 20 \,cm/s = 0.2 \,m/s$, Amplitude $A = 0.4 \,m$.First, calculate the angular

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$\sqrt{0.15} \,m$

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(B) Given: Mass $m = 1 \,kg$, Force constant $k = 4 \,N/m$, Velocity $v = 20 \,cm/s = 0.2 \,m/s$, Amplitude $A = 0.4 \,m$.First, calculate the angular frequency $\omega = \sqrt{k/m} = \sqrt{4/1} = 2 \,rad/s$.The velocity of a particle in $S.H.M.$ is given by $v = \omega \sqrt{A^2 - x^2}$, where $x$ is the displacement.Substituting the values: $0.2 = 2 \sqrt{0.4^2 - x^2}$.Dividing by $2$: $0.1 = \sqrt{0.16 - x^2}$.Squaring both sides: $0.01 = 0.16 - x^2$.Rearranging for $x^2$: $x^2 = 0.16 - 0.01 = 0.15$.Therefore, the displacement $x = \sqrt{0.15} \,m$.

$A$ spring executes $S.H.M.$ with a mass of $1 \,kg$ attached to it. The force constant of the spring is $4 \,N/m$. If at any instant its velocity is $20 \,cm/s$, the displacement at that instant is (Amplitude of $S.H.M.$ is $0.4 \,m$)

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