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=2π√m2K
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B
T=2π√mK
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C
T=2π√4mK
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D
T=2π√m4K
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Solution
The correct option is DT=2π√m4K 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=2π√μKeq
From (1) and (2), we can write that T=2π√m2(2K)⇒T=2π√m4K
Thus, option (d) is the correct answer.