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Question

Two masses m and 2m are suspended together by a massless spring of force constant K. When the masses are in equilibrium, mass m is removed without disturbing the system. Then, the amplitude of vibration of the system is:


A
2mgK
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B
mgK
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C
3mgK
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D
mg2K
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Solution

The correct option is B mgK

Let l be the extension of the spring at equilibrium, when both the masses are hanging.
From Hooke's law,
3mg=Kl(2)
Let l be the extension of the spring at equilibrium, when mass 2m alone is hanging.
From Hooke's law, we can write that
2mg=Kl(1)
This will be the new mean position for oscillation when mass m is removed.
Subtracting (1) from (2), we get
ll=mgK
i.e Amplitude of oscillation A=(ll)=mgK
Thus, option (b) is the correct answer.

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