Question

The transverse displacement of a string (clamped at its both ends) is given by

y(x, t) = 0.06 sin $(\frac{2π}{3 }x)$ cos (120 πt)

where x and y are in m and t in s. The length of the string is 1.5 m and its mass is 3.0 ×10–2 kg.

Answer the following :

(a) Does the function represent a travelling wave or a stationary wave?

(b) Interpret the wave as a superposition of two waves travelling in opposite directions. What is the wavelength, frequency, and speed of each wave?

(c) Determine the tension in the string

Answer 1

The general equation representing a stationary wave is given by the displacement function:

y (x, t) = 2a sin kx cos ωt

This equation is similar to the given equation:

$y(x,t)=0.06 \,sin\,(\frac{2}{3}x) cos(120 \pi\,t)$

Hence, the given function represents a stationary wave.

A wave travelling along the positive x-direction is given as:

y1=a sin(ωt - kx)

The wave travelling along the negative x-direction is given

y2=a sin(ωt - kx)

The superposition of these two waves yields:

y=y1+y2=a sin(ωt-kx)-a sin(ωt+kx)

=a sin(ωt) cos(kx) - a sin(kx) cos(ωt)-a sin(ωt) cos(kx)-a sin(kx) cos(ωt)

=-2a sin(kx) cos(ωt)

$=-2a \,sin(\frac{2\pi}{λ}x) cos(2 \,\pi\,vt)………(i)$

The transverse displacement of the string is given as:

y(x,t)=0.06 sin$(\frac{2\pi}{3}x)$ cos (120 $\pi$ t) …….(ii)

Comparing equations (i) and (ii), we have:

$\frac{2\pi}{λ}=\frac{2\pi}{3}$

∴Wavelength, λ = 3 m

It is given that:

120π = 2πν

Frequency, ν = 60 Hz

Wave speed, v = νλ

= 60 × 3 = 180 m/s

The velocity of a transverse wave travelling in a string is given by the relation:

$v=\sqrt\frac{T}{μ} ..........(i)$

Where,

Velocity of the transverse wave, v = 180 m/s

Mass of the string, m = 3.0 × 10–2 kg

Length of the string, l = 1.5 m

Mass per unit length of the string, $μ=\frac{m}{l}$

$=\frac{3.0}{1.5}×10^{-2}$

=2×1-2 kg m-1

Tension in the string = T

From equation (i), tension can be obtained as:

T = v 2 μ

= (180)2 × 2 × 10–2

= 648 N