The displacement of the string is given by,
$\mathrm{y}\left(\mathrm{x},\mathrm{t}\right)=0.06\mathrm{sin}\left(\frac{2\mathrm{\pi x}}{3}\right)\mathrm{cos}\left(120\mathrm{\pi t}\right)$
where x and y are in metres and t in seconds. The length of the string is 1.5 m and its mass is

(a) It represents a progressive wave of frequency 60 Hz.

(b) It represents a stationary wave of frequency 60 Hz.

(c) It is the result of a superposition of two waves of wavelength 3 m, frequency 60 Hz each travelling with a speed of 180 m/s in the opposite direction.

(d) The amplitude of this wave is constant.

Choose the correct alternatives:

1. (b, d)

2. (a, c)

3. (b, c)

4. (c, d)

(2) Hint: Use the standard equation of a stationary wave.
Given equation is, $\mathrm{y}\left(\mathrm{x},\mathrm{t}\right)=0.06\mathrm{sin}\left(\frac{2\mathrm{\pi x}}{3}\right)\mathrm{cos}\left(120\mathrm{\pi t}\right)$
Step 1: Find if it is a stationary wave.
1. Comparing it with a standard equation of stationary wave,

$\mathrm{y}\left(\mathrm{x},\mathrm{t}\right)=\mathrm{asin}\left(\mathrm{kx}\right)\mathrm{cos}\left(\mathrm{\omega t}\right)$
Clearly, the given equation belongs to the stationary wave. Hence, option 1 is not correct.
Step 2: Find the frequency of the wave.
2. By comparing;

Step 3: Find the wavelength of the wave.
3.
$\mathrm{k}=\frac{2\mathrm{\pi }}{3}=\frac{2\mathrm{\pi }}{\mathrm{\lambda }}$
Frequency= f = 60 Hz
Speed = v= $\mathrm{f\lambda }$ = (60 Hz) x (3m) = 180 m/s
4. In a stationary wave, all particles of the medium execute SHM with varying amplitude nodes.