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Question

Two blocks of masses m1 and m2 (m1>m2), are performing SHM together with the same amplitude and same time period as shown. Surface between m1 and the ground is smooth, while between m1 and m2, the coefficient of friction is μ. Given that k1m1>k2m2. Choose the correct option:


A
Time period of SHM is 2πm1m2(m1+m2)(k1+k2).
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B
Time period of SHM is 2π(m1+m2)(k1+k2)k1k2.
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C
Time period of SHM is 2πm1+m2k1+k2.
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D
Maximum possible amplitude of this SHM is 2μm2g(m1+m2)k1m2k2m1
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Solution

The correct option is C Time period of SHM is 2πm1+m2k1+k2.
Since the blocks perform SHM together (no slipping), we can take
Total mass, m=m1+m2 and keq=k1+k2
(springs in parallel)
Then, ω=k1+k2m1+m2
& T=2πm1+m2k1+k2

FBD of block m2 at extreme position (amplitude A):


We know, maximum acceleration of block m2
amax=ω2A=(k1+k2m1+m2)A

Direction of Friction is towards mean position as k1m1>k2m2

From FBD,
fmax+k2A=m2amax=m2(k1+k2m1+m2)A

For maximum possible amplitude, fmax=μm2g
μm2g=[m2(k1+k2m1+m2)k2]A
A=μm2g(m1+m2)k1m2k2m1

Hence, only option (c) is correct.

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