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Question

Two identical masses are connected to identical springs of spring constant K as shown in the figure. Find the time period of small oscillations.


A
T=2πm2K
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B
T=2πmK
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C
T=2π4mK
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D
T=2πm4K
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Solution

The correct option is D T=2πm4K
From the figure, we can deduce that the two springs are in parallel combination.
Keq=K1+K2=2K .....(1)
The diagram can be simplified as:


In the simplified diagram, two masses are connected by a spring of equivalent spring constant Keq=2K.
To reduce a two-body problem into a one-body problem, we use the concept of reduced mass.
μ=reduced mass=m1m2m1+m2
μ=(m)(m)m+m=m2 ....(2)
Time period of oscillations is given by,
T=2πμKeq
From (1) and (2), we can write that
T=2πm2(2K)T=2πm4K
Thus, option (d) is the correct answer.

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