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The given system is displacement by dist...

The given system is displacement by distance 'A' and released. Both the blocks (each of mass m) move together without relative slipping in the whole process. The magnetic of frictionless force between them at time 't' is

where `omega = sqrt((K)/(2m))`

A

`(KA)/(2)[cos omegat]`

B

`(KA)/(2)cos omegat`

C

`(KA)/(2)[sin omegat]`

D

`KA[cos omegat]`

Text Solution

Verified by Experts

The correct Answer is:
A
`a=omega^(2)x`
`a=(Kx)/(2m)=(KA)/(2m)cos omegat`
`f=ma=|(KA)/(2)cos omega t|`.
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