Question

A transverse wave is described by the equation $Y=Y_{0}sin2\pi (ft-\frac{x}{\lambda })$ . The maximum particle velocity is four times the wave velocity if

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

comparing the given equation with $y=asin(\omega t-kx)$, We get $a=Y_0,\omega =2\pi f,k=\frac{2\pi }{\lambda }.$
Hence maximum particle velocity $(v_{max}{)}_{particle}=aw=Y_0\times 2\pi f$ and wave velocity
$(v{)}_{wave}=\frac{w}{k}=\frac{2\pi f}{2\pi /\lambda }=f\lambda$
$\because (v_{max}{)}_{particle}=4v_{wave}\Rightarrow Y_0\times 2\pi f=4f\lambda \Rightarrow \lambda =\frac{\pi Y_0}{2}.$