Question

A travelling harmonic wave on a string is described by y(x, t) = 7.5 sin (0.005 x +12t + π /4) 

1. what are the displacement and velocity of oscillation of a point at x = 1 cm, and t = 1 s? Is this velocity equal to the velocity of wave propagation? 

2. Locate the points of the string which have the same transverse displacements and velocity as the x a 1 cm point at t = 2 s, 5 s, and 11s.

Answer 1

Given: 

y (x, t) = 7.5 sin (0.005 x + 12t + π /4)

1. At x=1 cm and t = 1s

y (1, 1)= 7.5 Sin(0.005 +12 + π /4)

= 7.5 sin (12.005 + π /4)

= 1.67 cm

Velocity of oscillation : v = \(\frac{d(Y (x,t))}{dt}\)

= d/dt (7.5 sin (0.005 x + 12t + π/4) dt

= 7.5 × 12 cos (0.005 x+ 12t + π/4)

At x = 1 cm and t = 1 s

v = 7.5 × 12 cos(0.005 + 12 + π/4)

= 87.75 cm s^-1

We know that velocity of wave propagation = w/k

Here w = 12 s^-1 and k = 0.005 cm^-1

∴ Velocity of wave propagation

= \(\frac{12s^{-1}}{0.005cm^{-1}}\) = 24 ms^-1

∴ At x = 1 cm and t = 1 s velocity of oscillation is not equal to velocity of wave propagation.

2. In a wave, all the points which are separated by a distance ± λ, ±2λ ……..from x = 1 cm will have same transverse displacements and velocity. For the given

wave , λ= \(\frac{2π}{0.005}\) = ±12.56 cm, +25.12

m….From x = 1 cm will have the same displacements and velocity as at x 

= 1 cm, t = 2s, 5s and 11 s.