Question

A travelling harmonic wave on a string is described by


(a) 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?

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

Answer 1

(a) Given equation ,
Y (x, t) = 7.5 sin( 0.005 x + 12t + π/4)

at x = 1 and t = 1 displacement of wave ( Y) = 7.5 sin( 0.00×1 + 12×1 + π/4)
= 7.5 sin( 12.79 rad )
= 7.5 sin ( 732.81°)
= 7.5 × 0.2217
= 1.6629 cm


Y = 7.5sin( 0.005x + 12t +π/4)
Differentiate wrt t
dY/dt = 12 × 7.5 cos(0.05x + 12t + π/4)
at t = 1 , x = 1

Velocity of particle (dy/dt) = 90cos(12.79)
= 90 × 0.975
= 87.75 cm /s


Now velocity of wave = w/K
= 12/0.005
= 2.4 × 10³ cm/s
= 24 m/s
Becoz sign of w and K are same so,
Velocity of wave = -24 m/s

Obviously you can see that
Velocity of particle is not same as velocity of wave .


(b) we know,
K = 2π/wavelength
Wavelength = 2π/K
= 2π/0.005 cm
= 12.56 m
All the points at n× wavelength ( where n = ±1 , ±2 , ±3 ..... ) away from x = 1cm will have same phase of motion . i.e.velocity and displacement
Hence, points are the 12.56 m, 25.12 m .....