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Question

A 1 kg block is executing simple harmonic motion of amplitude 0.1m 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 blocks move together, then


A
Amplitude of the motion becomes 5 cm
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B
The oscillation frequency becomes 52π Hz
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C
Amplitude of the motion becomes 10 cm
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D
The oscillation frequency becomes 5π Hz
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Solution

The correct options are
A Amplitude of the motion becomes 5 cm
B The oscillation frequency becomes 52π Hz
Given A=0.1m
k=100 N/m
m1=1 kg
m2=3kg

At the mean position, let 1 kg block has velocity v
Energy of system is conserved
12KA2=12m1v21
v1=KA2m1=100×0.121
v1=1 m/s
At the moment m1 is at mean position,
applying law of conservation of momentum,
m1v1=(m1+m2)v
1=(m1+m2)v
v=1(m1+m2)=14=0.25 m/s

New amplitude as 3 kg falls on 1 kg is
12KA'2=12(m1+m2)v2A2=(0.25)2×4100
A=5 cm

Frequency of the combined mass performing oscillation is given by
f=12πkm1+m2
f=12π1004
f=52π Hz
Thus, option (a) and (b) are correct.

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