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Question

A block of mass 0.9 kg attached to a spring of force constant k is lying on a friction less floor.The spring is compressed by 2 cm and the block is at a distance 12 cm from the wall as shown in the figure. When the block is released, it makes elastic collision with the wall and its period of motion is 0.2 sec. Find the appropriate value of k in Nm1
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Solution

x=Asin(ωt)
Let x=12, A=2, ω=2πT, T=0.2 sec
Time required to reach wall is given by
12=2sin(2πt0.2)
sin(10πt)=12t=160
solving this we get
t=T/6
Now total time of motion is
t=2×T/6
t=T/3
By problem t=0.2 sec
T=0.6 sec
T=2π×m/k
By solving we get K=100N/m

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