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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 π√mK+π√m2KThe mass m completes first half of oscillation (upward) under the effect of upper spring alone. Since, T=2π√mK, time taken for one half of oscillation t1=T2=π√mK 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=2π√m2K2=π√m2K Therefore, total time for 1 complete oscillation =t1+t2=π√mK+π√m2K

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