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Question

A 1 kg block is executing simple harmonic motion of amplitude 0.1 m on a smooth horizontal surface under the restoring force of a spring of spring constant 100 N/m. A block of mass 3 kg is gently placed on it at the instant it passes through the mean position. Assuming that the two blocks move together. Find the amplitude of the motion?

A
0.5m
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B
0.05m
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C
5m
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D
0.02m
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Solution

The correct option is B 0.05m
f=12πkmcombined mass=12π1004=2.5π=0.8 Hz
with 1 kg mass, f0=12π1001=5π Hz
Further, from conservation of linear momentum (at mean position)
ωA=14ω0A0
or fA=14f0A0
or A=f0A04f(5/π)(0.1)(4)(2.5/π)
=0.05m

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