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Question

A mass m fall on spring constant k and negligible mass from a height h. Assuming it sticks to the pan and executes simple harmonic motion, the maximum height upto which the pan will rise is:

A
mgk
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B
mgk[1+2khmg1]
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C
mgk[1+2khmg+1]
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D
mgk[1+khmg1]
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Solution

The correct option is C mgk[1+2khmg+1]

When the mass m falls on the spring from a height, let the length of the spring compressed be x.

Total distance by which the mass m falls is: (h+x)
Loss of PE of the mass is equal to the energy stored in the spring.
mg(h+x)=12kx2
x2(2mgk)x2mghk=0
x=2mgk±(2mgk)2+8mghk)2
x=mgk±(mgk)2+2mghk)=mgk(1+(1+2khmg))
We ignore the -ve sign, since it will give a negative x which is not a physical situation.
Thus, the maximum height to which the mass will rise is: mgk[1+(1+2khmg)]


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