Question

An object on the end of a spring with spring constant k moves in simple harmonic motion with amplitude A and frequency f. Which of th following is a possible expression for the kinetic energy of the object as a function of time t? (A) kA2 sin2 (2Tft) (B) AkA cos (2T ft) (C) kA sin (2n ft) (D) kA cos (2Tt ft)

Answer 1

Step-by-step explanation:

It is given that, an object on the end of a spring with spring constant k moves in simple harmonic motion with amplitude A and frequency f.

The equation of a particle executing SHM is given by :

`x=A\ sin\omega t`.........(1)

Where

A is the amplitude of the wave

Differentiating equation (1) wrt t as :

`v=(dx)/(dt)=(d(A\ sin\omega t))/(dt)`

`v=A\omega\ cos\omega t`.........(2)

The kinetic energy of the particle is given by :

`K=(1)/(2)mv^2`

`K=(1)/(2)m(A\omega\ cos\omega t)^2`

`K=(1)/(2)mA^2\omega^2\ cos^2\omega t`.........(3)

We know that,

`\omega=\sqrt{(k)/(m)}`

So, equation (3) becomes :

`K=(1)/(2)kA^2cos^2\omega t`

or

`K=(1)/(2)kA^2cos^2(2\pi ft)`

So, the kinetic energy of the object is
`(1)/(2)kA^2cos^2(2\pi ft)`. Hence, this is the required solution.