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Question

A block of mass 10 kg attached to a horizontal spring with force constant 250 N/m is moving with simple harmonic motion having amplitude 10 m. At the instant when the block passes through its equilibrium position, a lump of putty with mass 5 kg is dropped vertically on the block from a very small height and sticks to it. Find the new amplitude and period?

A
8.1 m,1.54 s
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B
10 m,1.54 s
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C
8.1 m,1.25 s
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D
12.2 m,1.54 s
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Solution

The correct option is A 8.1 m,1.54 s
Given,
mass of block (M)=10 kg
mass of pulley (m)=5 kg
Amplitude (A1)=10 m
Force constant (K)=250 N/m

Time period of oscillation as the putty drops on to the mass
(T2)=2πM+mK=2×3.14×15250=1.54 s

Let v1 be the speed of block at equilibrium position,
v2 be the speed of the combined mass,
Momentum of the system in horizontal direction is conserved, then
(M+m)v2=Mv1
v2=(MM+m)v1 ...(2)
Now, let A2 be the amplitude of system afterwards, then
A2ω2=(MM+m)A1ω1
A2=(MM+m)A1ω1ω2
A2=(MM+m)A1M+mM
A2=A1MM+m
=10×1015=1023=8.1 m

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