The velocity `v` of a particle of mass is moving along a straight line change with time `t` as `(d^(2)v)/(dt^(2)) = - Kv` where `K` is a particle constant which of the following statement is correct?

Options

A. the particle does not perform SHM.
B. The particle performs SHM with time period `2pisqrt((m)/(K))`
C. The particle performs SHM with frequency `(sqrtK)/(2pi)`
D. The particle performs SHM with time period `(2pi)/(K)`.

Answer

C

Explanation

`(d^2v)/(dt)=-kv^2`
This equation has standard solution `v=v_0sin(sqrtkt+theta)`, where `omega=sqrtk`
hence the particle executes SHM with angular frequency `omega=sqrtk`
or frequency `f=(sqrtk)/(2pi)`