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If a body is executing simple harmonic m...

If a body is executing simple harmonic motion and its current displacement is `sqrt(3)//2` times the amplitude from its mean position , then the ratio between potential energy and kinetic energy is

A

`3:2`

B

`2:3`

C

`sqrt(3):1`

D

`3:1`

Text Solution

Verified by Experts

The correct Answer is:
D

(d) The current displacement is `(sqrt(3))/(2)` times the amplitude
i.e., `y=(sqrt(3))/(2)A`
we know that potential energy of a body in SHM is given by where, m and `omega` are constant.
`U=(1)/(2)momega^(2)y^(2)`
`U=(1)/(2)momega^(2)((sqrt(3))/(2)A)^(2)=(1)/(2)momega^(2)xx(3A^(2))/(4)" "...(i)`
Similarly
Kinetic energy of a body in SHM is given by
`K=(1)/(2)momega^(2)(A^(2)-y^(2))=(1)/(2)momega^(2)(A^(2)-(3A^(2))/(4))=(1)/(2)momega^(2)(A^(2))/(4)" "...(ii)`
So, the ratio of PE and KE
`PE: KE =(1)/(2)momega^(2)xx(3A^(2))/(4):(1)/(2)momega^(2)(A^(2))/(4)=3:1`
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