The transverse displacement of a string (clamped at its both ends) is given by \(y(x,t)=0.6 sin\left ( \frac{2\pi }{3} x\right )cos (120 \pi t)\) where x and y are in m and t in s. The length of the string is 1.5m and its mass is \(3.0\times 10^{-2} kg\).
Answer the following:
(a) Does the function represent a traveling wave or a stationary wave?
(b) Interpret the wave as a superposition of two waves traveling in opposite directions. What is the wavelength, frequency, and speed of each wave?
(c) Determine the tension in the string.

The general equation representing a stationary wave:

y (x, t) = 2asinkxcos ωt
This equation is similar to the equation:
y(x, t)=0.06 sin2π3xcos (120πt)
Hence, the given function represents a stationary wave.
A wave travelling along the positive x-direction is:
y1=a sin(ωt-kx)

The wave travelling along the negative x-direction is:
y2=a sin(ωt+kx)

The superposition of these two waves yields:

y=y1+y2=asin(ωtkx)asin(ωt+kx)=asinωtcoskxasinkxcosωtasinwtcoskxasinkxcoswt=2asinkxcosωt=2asin2πλxcos2πνt                      ...i

The transverse displacement of the string is:

y(x, t)=0.06sin2π3xcos(120 πt)               ...ii

Comparing equations (i) and (ii),

Wavelength, λ=3 m 
Frequency, ν=60 Hz
 Wave speed, v=νλ=60×3=180 m/s 
The velocity of a transverse wave travelling in a string:
v=Tμ                 ...i
Velocity of the transverse wave, v = 180 m/s 
Mass of the string, m = 3.0 × 102 kg
Length of the string, l = 1.5 m 
Mass per unit length of the string,  μ=ml=3.01.5×10-2=2×10-2 kg m-1
Tension in the the string, 
T = v2μ 
= (180)2 × 2 × 10-2
 = 648 N