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The ration of kinetic energy to the pote...

The ration of kinetic energy to the potential energy of a particle executing SHM at a distance equal to half its amplitude , the distance being measured from its equilibrium position is

A

`2:1`

B

`3:1`

C

`8:1`

D

`1:1`

Text Solution

Verified by Experts

The correct Answer is:
B
(b) As, potential energy of a particle executing simple harmonic motion also periodic with period `T//2` .So, potential extreme displacements.
Let the amplitude of SHM be A.
Now, potential energy of SHM `=(1)/(2)kx^(2)`
Here, `x=(A)/(2)impliesU=(1)/(2)k(A^(2))/(4)" "...(i)`
Kinetic energy , `K=(1)/(2)kA^(2)-(1)/(2)k(A^(2))/(4)impliesK=(3)/(8)kA^(2)" "...(ii)`
On dividing Eq. (ii) by Eq. (i) , we get
`(K)/(U)=(3)/(8)xx(8)/(1)=(K)/(U)=(3)/(1)`
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