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 $t=\frac{T}{12}$