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Question

One end of a light spring of force constant K is fixed to a block of mass M placed on a horizontal frictionless surface, the other end of the spring being fixed to a wall. The spring – block system is executing simple harmonic motion of amplitude A and frequency v. When the block is passing through the equilibrium position, an object of a mass m is gently placed on the block. As a result, the frequency of the system becomes v' and the amplitude becomes A' . The ratio v'/v will be,


A

(MM+m)1/2

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B

(mM+m)1/2

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C

MAmA

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D

[(M+m)AmA]

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Solution

The correct option is A

(MM+m)1/2


The frequency of the system before the object is placed on the block is given by

v=12π(kM+m)1/2

After the object of mass m is placed on the block, the new frequency of the system becomes

v=12π[kM+m]1/2

vv=(MM+m)1/2


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