### Q.22:- 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:-**

(a) The given harmonic wave is

= 90 coss (732.81°) = 90 cos (90 × 8 + 12.81°)

= 90 cos (12.81°)

= 90 × 0.975 =87.75 cm/s

Now, the equation of a propagating wave is given by:

*y*(

*x*,

*t*) =

*a*sin (

*kx*+

*wt*+

*Φ*)

where,

*k* = 2π/*λ*

∴ λ = 2π/*k*

and, ω = 2π**v**

∴ **v** = ω/2π

Speed, *v* = * vλ = *ω/

*k*

where,

ω = 12rad/s

*k*= 0.0050 m

^{-1}

∴

*v*= 12/0.0050 = 2400 cm/s

Hence, the velocity of the wave oscillation at x = 1 cm and t = 1s is not equal to the velocity of the wave propagation.

(b) Propagation constant is related to wavelength as:

k = 2π/*λ*

∴ *λ = *2π/*k *= 2 × 3.14 / 0.0050

= 1256 cm = 12.56 m

Therefore, all the points at distance *nλ* (*n* = ±1, ±2, …. and so on), i.e. ± 12.56 m, ± 25.12 m, … and so on for *x* = 1 cm, will have the same displacement as the *x* = 1 cm points at* t* = 2 s, 5 s, and 11 s.