Question

Two pendulum oscillate with a constant phase difference of $45^{\circ}$ and same amplitude. If the maximum velocity of one of them is $v$ and that of other is $v+x$ then the value of $x$ will be

• A. $0$
• B. $v/2$
• C. $v/\sqrt{2}$
• D. $(\sqrt{2})v$

Answer 1

Answer: (A)

$0$

Answer 2

Let velocity of one pendulum is
$v_{1}=v_{0} \cos \left(\omega t+\phi_{1}\right)$
Velocity of the other pendulum is
$v_{2}=v_{0} \cos \left(\omega t+\phi_{2}\right)$
According to the question,
$\phi_{2}-\phi_{1}=45^{o}$
Clearly, $\left|v_{1}\right|_{\max }=\left|v_{2}\right|_{\max }=v_{0}=v$
Thus, $x=0$