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A clock is performing SHM along a vertic...

A clock is performing SHM along a vertical line with amplitude of 40 cm on a horizontal plank. The block just lose the contact with plank when plank is momentarily at rest. Then, (Take, `g=10m//s^(2)`)

A

the period of its oscillation is `(2pi)/(5)s`

B

the period of is oscillation is `(2pi)/(6)s`

C

the period of its oscillation is `(pi)/(5)s`

D

None of the above

Text Solution

Verified by Experts

The correct Answer is:
A
The block just loose the contact from the plank if acceleration of plank just equal to g. (Vertiaclly downward)
Since, block just loses contact, when the plank is momentarily at rest. Therefore, at that instant the plank is at its upper extreme postion.
But at extreme position acceleration `=aomega^(2)`
where, a = 40 cm = `0.4`m
Hence, `0.4omega^(2)=g=10m//s^(2)`
`omega=5rad//s" "rArr" "T=(2pi)/(5)s`
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