A wooden block is suspended from a spring. Its period for vertical oscillation is T. In a second experiment the same apparatus is laid on the horizontal frictionless table with the free-end of the spring fixed to a nail on the table. The block is pulled a little so as to stretch the spring and then released. The mass will

execute SHM with period T

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execute SHM with period T/2

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execute SHM with period 2T

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does not execute SHM

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Solution

The correct option is A execute SHM with period T
The natural frequency of the system is $ω = \sqrt{k / m} \Rightarrow T = 2 π \sqrt{m / k}$ . where $m$ is the mass of the block and $k$ is the spring constant.
In both the cases, the mass will execute SHM with the time period $T = 2 π \sqrt{m / k}$

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The correct option is A execute SHM with period TThe natural frequency of the system is ω=√k/m⇒T=2π√m/k. where m is the mass of the block and k is the spring constant.In both the cases, the mass will execute SHM with the time period T=2π√m/k