Answer to Question #117966 in Mechanics | Relativity for sridhar

Question

A body is in simple harmonic motion with time period T=0.5s and amplitude A=1cm. Find the average velocity in the interval in which it moves from equilibrium position to half of its amplitude.
Ans 12cm/s

Answer 1

Given that, a body in simple harmonic motion with time period $T=0.5s$ and and amplitude is $A=1cm=10^{-2}m$ .

The equation of simple harmonic motion will be,

$$y=A\sin(\frac{2\pi}{T}t)\hspace{1cm}(\spades)$$

Let at $t=t_1$ ,the position of the body from equilibrium is $\frac{A}{2}$ ,thus from above equation we get,

$$\frac{A}{2}=A\sin(\frac{2\pi}{T}t_1)\\ \implies \sin(\frac{2\pi}{T}t_1)=\frac{1}{2}\\ \implies \frac{2\pi}{T}t_1=\frac{\pi}{6}\\ \implies t_1=\frac{T}{12}\hspace{1cm}(\clubs)$$

Clearly, if we differentiate $(\spades)$ w.r.t time we get the expression for velocity of the body,thus

$$v=\frac{2\pi A}{T}\cos(\frac{2\pi }{T}t)$$

Hence,

$$<v>=\frac{1}{t_1-0}\int_{0}^{t_1}vdt\\ \implies <v>=\frac{1}{t_1-0}\int_{0}^{t_1}\frac{2\pi A}{T}\cos(\frac{2\pi }{T}t)dt\\ \implies <v>=\frac{2\pi A}{Tt_1}\int_{0}^{t_1}\cos(\frac{2\pi }{T}t)dt\\ \implies <v>=\frac{2\pi A}{Tt_1}\frac{T}{2\pi}\bigg[\sin(\frac{2\pi }{T}t)\bigg]_{0}^{t_1}\\ \implies <v>=\frac{A}{t_1}\sin(\frac{\pi }{6})=\frac{A}{2t_1}=12cm/s$$