The displacement of a string is given by `y(x,t)=0.06sin(2pix//3)cos(120pit)` where x and y are in m and t in s. The lengthe of the string is 1.5m and its mass is `3.0xx10^(-2)kg.`

Options

A. It represents a progresssive wave of frquency 60Hz
B. It represens a stationary wave of frequency 60Hz
C. It is the result superpositon of two waves of wavelength 3m , frequency 60 Hz each travelling with a speed of 180 m/s in ipposite direction
D. Amplitude of this wave is constant

Explanation

To solve the problem, we will analyze the given wave equation and determine the characteristics of the wave, including whether it is a progressive wave or a stationary wave, its frequency, wavelength, and amplitude.

### Step-by-Step Solution:

1. **Identify the Wave Equation:**
The displacement of the string is given by:
\[ y(x,t) = 0.06 \sin\left(\frac{2\pi x}{3}\right) \cos(120\pi t) \]

2. **Recognize the Form of the Wave Equation:**
The wave equation can be compared to the general form of a stationary wave:
\[ y(x,t) = A \sin(kx) \cos(\omega t) \]
Here, \(A\) is the amplitude, \(k\) is the wave number, and \(\omega\) is the angular frequency.

3. **Determine the Amplitude:**
From the equation, the amplitude \(A\) is:
\[ A = 0.06 \, \text{m} \]

4. **Find the Wave Number \(k\):**
The wave number \(k\) is given by:
\[ k = \frac{2\pi}{\lambda} \]
From the equation, we see:
\[ k = \frac{2\pi}{3} \implies \lambda = 3 \, \text{m} \]

5. **Determine the Angular Frequency \(\omega\):**
The angular frequency \(\omega\) is:
\[ \omega = 120\pi \, \text{rad/s} \]

6. **Calculate the Frequency \(f\):**
The frequency \(f\) can be calculated using the relation:
\[ \omega = 2\pi f \implies f = \frac{\omega}{2\pi} = \frac{120\pi}{2\pi} = 60 \, \text{Hz} \]

7. **Identify the Type of Wave:**
Since the wave equation is of the form \(y(x,t) = A \sin(kx) \cos(\omega t)\), it represents a stationary wave, not a progressive wave.

8. **Check for Superposition of Waves:**
A stationary wave is formed by the superposition of two waves traveling in opposite directions. The component waves can be expressed as:
\[ y_1 = A \sin(kx - \omega t) \quad \text{and} \quad y_2 = A \sin(kx + \omega t) \]
The resultant wave is indeed a result of the superposition of two waves.

9. **Amplitude Variation:**
The amplitude of the standing wave varies with position \(x\) as:
\[ A(x) = 0.06 \sin\left(\frac{2\pi x}{3}\right) \]
This indicates that the amplitude is not constant but varies with position.

### Conclusion:
- The wave is a stationary wave with a frequency of 60 Hz.
- It is the result of the superposition of two waves with a wavelength of 3 m.
- The amplitude of the wave is not constant; it varies with position.