Question

Consider a sinusoidal travelling wave shown in figure. The wave velocity is $+40cm/s$. The velocity of a particle at point P at the instant shown is $-126\times {10}^{-x}m/s$. Find $x$.

• A. $5$
  
• B. $2$
  
• C. $4$
  
• D. $6$
  

Answer 1

Answer: (B)

$2$

We have $v=\lambda \nu$

$⟹\nu =\frac{v}{\lambda }=\frac{40cm/s}{4cm}=10Hz$

The wave must have a form $y=Asin(kx-\omega t+\varphi )$

Thus the particle velocity is given as $v=-A\omega cos(kx-\omega t+\varphi )$

At point P, the $sin(\omega t-kx+\varphi )$ term vanishes.

Thus $v=-A\omega =-2cm\times 2\pi \nu =-2cm\times 2\pi \times 10Hz=-126cm/s=-126\times {10}^{-2}m/s$

Therefore $x=2$

The negative sign comes from the realization that the wave is right travelling. Thus after an instant, each point on the wave would have a displacement which the point instantly left to it had a moment ago. Thus the point P would move in the negative y direction.