The velocity `v` of a particle of mass in moving along a straight line change within 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 perform `SHM`with time period `2pi sqrt((m)/(k))`
C. The particle perform `SHM`with time frequency`sqrt(k)/(2pi)`
D. The particle perform `SHM`with time period `(2pi)/(K)`

Answer

C

Explanation

`(d^(2)v)/(dt) = - kv^(2)`
this equation has standard solution `v = v_(0) sin(sqrt(k)t + theta)`where `= omega = sqrt(k)`
Hence the particle executes `SHM` with angular frequency `omega = sqrt(k) ` or frequency `f = sqrt(k)/(2pi)`.