Question

A particle of mass ‘m’ is executing S.H.M. about the origin on x-axis with frequencу `sqrt ((k a)/(pi m))`, where ‘k’ is a constant and ‘a’ is the amplitude of S.H.M. If ‘x’ is a displacement of a particle, at time ‘t’, potential energy of the particle will be ______.

पर्याय

• `1/2 kax^2`

• πkax²

• 2πkax²

• 2kax²

Answer 1

A particle of mass ‘m’ is executing S.H.M. about the origin on x-axis with frequencу `sqrt ((k a)/(pi m))`, where ‘k’ is a constant and ‘a’ is the amplitude of S.H.M. If ‘x’ is a displacement of a particle, at time ‘t’, potential energy of the particle will be 2πkax².

Explanation:

Given: Frequency (f) = `sqrt (k a)/(pi m)`

For S.H.M., potential energy:

U = `1/2 m omega^2 x^2`

Angular frequency (ω) = 2πf

= `2 pi sqrt ((k a)/(pi m)`

ω² = `(2 pi)^2 * (k a)/(pi m)`

= `4 pi^2 * (k a)/(pi m)`

= `(4 pi k a)/m`

Substituting these values in equation (i), we get,

U = `1/2 m * (4 pi k a)/m * x^2`

= 2πkax²