Question

The stationary wave in a vibrating wire of sonometer is represented by an equation $y=2A\sin \left(kx\right)\cos \left(200t\right)$ where all physical quantities in the equation are in S.I units. If velocity of particle of wire at antinode is $10m/s$, the amplitude of particle at antinode is

• A. $0.04m$
• B. $0.05m$
• C. $0.08m$
• D. $0.1m$

Answer 1

The correct option is B $0.05m$

particle velocity $v=\frac{dy}{dt}=400A\sin kx\sin 200t$
Amplitude of particle at antinode is $2A$
since $\sin kx=1$ at antinode
$\therefore 400A=10$
$2A=0.05m$