The transverse displacement of a string fixed at both ends is given by `y=0.06sin((2pix)/(3))cos(100pit)` where x and y are in metres and t is in seconds. The length of the string is 1.5 m and its mass is `3.0xx10^(-2)kg`. What is the tension in the string?

Options

A. 225 N
B. 300 N
C. 450 N
D. 675 N

Explanation

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

### Step-by-Step Solution:

1. **Identify the given wave equation**: The transverse displacement of the string is given by:
\[ y = 0.06 \sin\left(\frac{2\pi x}{3}\right) \cos(100\pi t) \]
Here, \(x\) and \(y\) are in meters, and \(t\) is in seconds.

2. **Extract the wave parameters**: From the wave equation, we can identify:
- The wave number \(k\) is given by the coefficient of \(x\) in the sine function:
\[ k = \frac{2\pi}{3} \, \text{m}^{-1} \]
- The angular frequency \(\omega\) is given by the coefficient of \(t\) in the cosine function:
\[ \omega = 100\pi \, \text{rad/s} \]

3. **Calculate the wave speed \(v\)**: The wave speed can be calculated using the formula:
\[ v = \frac{\omega}{k} \]
Substituting the values we found:
\[ v = \frac{100\pi}{\frac{2\pi}{3}} = \frac{100\pi \cdot 3}{2\pi} = 150 \, \text{m/s} \]

4. **Calculate mass per unit length \(\mu\)**: The mass per unit length \(\mu\) of the string can be calculated using:
\[ \mu = \frac{m}{L} \]
where \(m = 3.0 \times 10^{-2} \, \text{kg}\) and \(L = 1.5 \, \text{m}\):
\[ \mu = \frac{3.0 \times 10^{-2}}{1.5} = 2.0 \times 10^{-2} \, \text{kg/m} \]

5. **Use the wave speed to find tension \(T\)**: The relationship between wave speed, tension, and mass per unit length is given by:
\[ v = \sqrt{\frac{T}{\mu}} \]
Rearranging this formula gives:
\[ T = v^2 \cdot \mu \]
Substituting the values we calculated:
\[ T = (150)^2 \cdot (2.0 \times 10^{-2}) = 22500 \cdot 2.0 \times 10^{-2} = 450 \, \text{N} \]

### Final Answer:
The tension in the string is \(450 \, \text{N}\).

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