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Question

A mass m is suspended from a spring of spring constant K and just touches another identical spring fixed to the floor as shown in the figure. The time period of small oscillations is


A
2πmK
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B
πmK+π mK2
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C
π m3K2
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D
πmK+πm2K
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Solution

The correct option is D πmK+πm2K
The mass m completes first half of oscillation (upward) under the effect of upper spring alone.
Since, T=2πmK, time taken for one half of oscillation t1=T2=πmK
When the mass goes downwards, it is under the combined action of both the springs.
Both springs are in parallel combination.
So, Keq=K1+K2=2K
Time taken to complete second half of oscillation
t2=T2=2πm2K2=πm2K
Therefore, total time for 1 complete oscillation
=t1+t2=πmK+πm2K

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