Question

In S.H.M, the loss of kinetic energy is proportional to:
$\text{A.}\quad e^x$
$\text{B.}\quad x^3$
$\text{C.}\quad x^2$
$\text{D.}\quad x$

Answer 1

A motion in which a particle undergoes periodic motion is called Simple harmonic motion (S.H.M). Not every periodic motion is S.H.M but every S.H.M is periodic motion. In order to get the relation between kinetic energy and displacement (by seeing the options), first we need to get the expression of velocity by differentiating the displacement expression (i.e. the equation of S.H.M).

Formula used:
$|v|=\dfrac{dx}{dt}, K.E = \dfrac12 mv^2$

Complete step-by-step answer:
First, let’s understand the standard wave equation.
$Y=asin\left( \omega t+\phi \right)$is called the standard wave equation.
Here ‘Y’ represents the displacement of wave particles at time ‘t’. Coefficient of trigonometric function ‘a’ is called the amplitude of the wave. ’$\omega$’ is the angular frequency of the wave, which is the measure of angular displacement. ‘$\phi$’ is the initial phase difference of the wave. It is also called ‘epoch’.
Now, velocity is $\dfrac{dY}{dt}$, as here, position is represented by ‘Y’, as velocity is the rate of change of displacement.
So, differentiating the above equations, we get:
${{v}}=a \omega cos\left(\omega t+\phi \right)$ $\left[ \dfrac{d(sin(ax+b)}{dx}=acos(ax+b) \right]$
Now, we can write velocity equation as:
$cos(\omega t +\phi) = \dfrac{v}{a\omega}$
And displacement term as:
$sin(\omega t +\phi) = \dfrac{Y}{a}$
Now, as $sin^2\theta + cos^2\theta = 1$
Thus, $cos^2(\omega t +\phi) +sin^2(\omega t +\phi) = \left(\dfrac{v}{a\omega}\right)^2 + \left(\dfrac{Y}{a}\right)^2 =1$
Now, say Y =x (displacement), so
$\left(\dfrac{v}{a\omega}\right)^2 + \left(\dfrac{x}{a}\right)^2 =1$
Now, $K.E. = \dfrac12 mv^2$
Thus the loss of K.E. is proportional to $v^2$ and from $\left(\dfrac{v}{a\omega}\right)^2 + \left(\dfrac{x}{a}\right)^2 =1$, we can say that $v^2$ varies as a function of $x^2$, so option C. is correct.

So, the correct answer is “Option C”.

Note: In S.H.M, the loss of energy is taken zero as otherwise it’ll become damped harmonic motion D.H.M. But then why in the question, we’ve been asked for the loss of Kinetic energy? This is because the kinetic energy gets converted to potential energy and this conversion is cyclic in nature which gets continued without the actual loss of energy.