Question

At time $t=0$, one particle is at maximum positive amplitude and the other particle is at half of the positive amplitude. Their amplitudes and time periods $T$ are same. If they are approaching, find the time by which they cross each other:

• A. $\frac{T}{6}$
• B. $\frac{T}{12}$
• C. $\frac{T}{16}$
• D. $\frac{T}{24}$

Answer 1

Answer: (B)

$x_1=A\cos \omega t\left(\text{ at }t=0,x_1=A\right)$
$x_2=A\sin \left(\omega t+\frac{\pi }{6}\right)\left(\text{ at }t=0,x_2=+A/2\right)$
For $x_1=x_2$,
$\cos \omega t=\sin (\omega t+\pi /6)$
$=\sin \omega t\cos \frac{\pi }{6}+\cos \omega t\sin \frac{\pi }{6}$
$\cos \omega t\left(1-\sin \frac{\pi }{6}\right)=\sin \omega t\cos \frac{\pi }{6}$
$\tan \omega t=\frac{(1-\sin \pi /6)}{\cos \pi /6}=\frac{1}{\sqrt{3}}$
$\omega t=\left(\frac{2\pi }{T}\right)t=\frac{\pi }{6}$
Will give $t=\frac{T}{12}$