Question

The amplitude of wave disturbance propagating in the positive x-axis is given by y = $\frac{1}{x^{2}–2x + 1}$ at

t = 2 sec and y = $\frac{1}{x^{2} + 2x + 5}$ at t = 6 sec, where x and y are in meters. Velocity of the pulse is –

• A. 1 m/s in positive x-direction
• B.
  • 2 m/s in negative x-direction
• C. 0.5 m/s in negative x-direction
• D. 1 m/s in negative x-direction

Answer 1

Answer: (C)

0.5 m/s in negative x-direction

Answer 2

At t = 2 sec,

y =$\frac{1}{x^{2}–2x + 1}$

At t = 6 sec,

y = $\frac{1}{x^{2} + 2x + 5}$

Ž y = $\frac{1}{(x + 2)^{2}–2x + 1}$

\ Wave velocity = $\frac{2}{4}$ = $\frac{1}{2}$ m/s in negative x-direction.