The transverse displacement of a string clamped at its both ends is given by
`y(x, t) = 0.06 sin ((2pi)/3 x) cos(l20pit)` 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 xx 10^(-2)` kg. The tension in the string is

Options

A. 324 N
B. 648 N
C. 832 N
D. 972 N

Explanation

To find the tension in the string given the transverse displacement equation, we can follow these steps:

### Step 1: Identify the parameters from the wave equation
The given wave equation is:
\[ y(x, t) = 0.06 \sin\left(\frac{2\pi}{3} x\right) \cos(120\pi t) \]

From this equation, we can identify:
- Amplitude \( a = 0.06 \, \text{m} \)
- Angular frequency \( \omega = 120\pi \, \text{rad/s} \)
- Wave number \( k = \frac{2\pi}{3} \, \text{rad/m} \)

### Step 2: Calculate the frequency \( \nu \)
The relationship between angular frequency \( \omega \) and frequency \( \nu \) is given by:
\[ \omega = 2\pi\nu \]
Thus,
\[ \nu = \frac{\omega}{2\pi} = \frac{120\pi}{2\pi} = 60 \, \text{Hz} \]

### Step 3: Calculate the wavelength \( \lambda \)
The wave number \( k \) is related to the wavelength \( \lambda \) by:
\[ k = \frac{2\pi}{\lambda} \]
Rearranging gives:
\[ \lambda = \frac{2\pi}{k} = \frac{2\pi}{\frac{2\pi}{3}} = 3 \, \text{m} \]

### Step 4: Calculate the wave speed \( v \)
The speed of the wave \( v \) is given by the product of frequency and wavelength:
\[ v = \nu \lambda = 60 \times 3 = 180 \, \text{m/s} \]

### Step 5: Calculate the mass per unit length \( m \)
The mass of the string is given as \( 3 \times 10^{-2} \, \text{kg} \) and the length of the string is \( 1.5 \, \text{m} \). Therefore, the mass per unit length \( m \) is:
\[ m = \frac{\text{mass}}{\text{length}} = \frac{3 \times 10^{-2}}{1.5} = 2 \times 10^{-2} \, \text{kg/m} \]

### Step 6: Use the wave speed to find the tension \( T \)
The relationship between wave speed \( v \), tension \( T \), and mass per unit length \( m \) is given by:
\[ v = \sqrt{\frac{T}{m}} \]
Squaring both sides gives:
\[ v^2 = \frac{T}{m} \]
Rearranging for tension \( T \):
\[ T = m v^2 \]

Substituting the values:
\[ T = (2 \times 10^{-2}) \times (180)^2 \]
\[ T = (2 \times 10^{-2}) \times 32400 \]
\[ T = 648 \, \text{N} \]

### Final Answer
The tension in the string is \( 648 \, \text{N} \).

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