A body of mass 1 ,kg is suspended from a spri | vedclass

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(D) Given: Mass of suspended body $M = 1 \,kg$,mass of hitting body $m = 0.5 \,kg$,velocity of hitting body $v = 3 \,ms^{-1}$,and frequency $f = \frac

Vedclass · Accepted Answer

$5 \,cm, 600 \,Nm^{-1}$

Vedclass · Answer

(D) Given: Mass of suspended body $M = 1 \,kg$,mass of hitting body $m = 0.5 \,kg$,velocity of hitting body $v = 3 \,ms^{-1}$,and frequency $f = \frac{10}{\pi} \,Hz$.After collision,the total mass of the system is $M_{total} = M + m = 1 + 0.5 = 1.5 \,kg$.The frequency of oscillation is given by $f = \frac{1}{2\pi} \sqrt{\frac{k}{M_{total}}}$.Rearranging for the spring constant $k$: $k = (2\pi f)^2 M_{total} = (2\pi 	imes \frac{10}{\pi})^2 	imes 1.5 = (20)^2 	imes 1.5 = 400 	imes 1.5 = 600 \,Nm^{-1}$.Using conservation of linear momentum during the collision: $m v = (M + m) v'$,where $v'$ is the velocity of the combined mass immediately after collision.$0.5 	imes 3 = 1.5 	imes v' \Rightarrow 1.5 = 1.5 v' \Rightarrow v' = 1 \,ms^{-1}$.The amplitude $A$ is related to the maximum velocity $v'$ by $v' = A \omega$,where $\omega = 2\pi f = 2\pi 	imes \frac{10}{\pi} = 20 \,rad/s$.$A = \frac{v'}{\omega} = \frac{1}{20} \,m = 0.05 \,m = 5 \,cm$.Thus,the amplitude is $5 \,cm$ and the spring constant is $600 \,Nm^{-1}$.

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