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