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

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 :

$v=\dfrac{dx}{dt}=\dfrac{d(A\ sin\omega t)}{dt}$

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

The kinetic energy of the particle is given by :

$K=\dfrac{1}{2}mv^2$

$K=\dfrac{1}{2}m(A\omega\ cos\omega t)^2$

$K=\dfrac{1}{2}mA^2\omega^2\ cos^2\omega t$.........(3)

We know that,

$\omega=\sqrt{\dfrac{k}{m}}$

So, equation (3) becomes :

$K=\dfrac{1}{2}kA^2cos^2\omega t$

or

$K=\dfrac{1}{2}kA^2cos^2(2\pi ft)$

So, the kinetic energy of the object is $\dfrac{1}{2}kA^2cos^2(2\pi ft)$. Hence, this is the required solution.