Question

A particle performs SHM with a period T and amplitude a.The mean velocity of the particle over the time interval during which it travels a distance a/2 from the extreme position is

• A. a/T
• B. 2a/T
• C. 3a/T
• D. a/2T

Answer 1

Answer: (C)

3a/T

Since the particle is starting from extreme position. We can take its equation as $x=a\cos \omega t$
at $t=t_1$ particle is at extreme position, i.e. $x=a$, we have
$a=a\cos \omega t_1$
$\Rightarrow \cos \omega t_1=1\Rightarrow \omega t_1=0\Rightarrow t_1=0$
at $t=t_2$ particle is at a distance $\frac{a}{2}$ from extreme position, i.e. $x=\frac{a}{2}$, we have
$\frac{a}{2}=a\cos \omega t_2$
$\Rightarrow \cos \omega t_2=\frac{1}{2}\Rightarrow \omega t_2=\frac{\pi }{3}\Rightarrow t_2=\frac{\pi }{3\omega }$
$⟨v⟩=$total distance/total time
$\Rightarrow ⟨v⟩=\frac{\left(\frac{a}{2}\right)}{t_2-t_1}=\frac{\left(\frac{a}{2}\right)}{\frac{\pi }{3\omega }-0}=\frac{3a\omega }{2\pi }=\frac{3a}{2\pi }\left(\frac{2\pi }{T}\right)=\frac{3a}{T}$