Question

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

[IIT 1984; MP PMT 1997; EAMCET; 1998;CBSE PMT 2000; AFMC 2000; MP PMT/PET 1998; 01;KCET 1999, 04; Pb. PET 2001; DPMT 2005]

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

Comparing the given equation with $y=asin(\omega t-kx)$, We get $a=Y_o$, $\omega =2\pi f$, $k=\frac{2\pi }{\lambda }$. Hence maximum particle

velocity $(v_{max}{)}_{particle}=a\omega =Y_o\times 2\pi f$ and wave velocity $(v{)}_{wave}=\frac{w}{k}=\frac{2\pi f}{2\pi /\lambda }=f\lambda$

$\because (v_{max}{)}_{particle}=4v_{wave}\Rightarrow Y_o\times 2\pi f=4f\lambda \Rightarrow \lambda =\frac{\pi Y_0}{2}$