Equation of travelling wave on a stretched string of linear density `4 kg m^(-1)` is `y = 0.07 cos(250t - 15x)`, where distance and time are measured in SI units. The tension in the string be xN. Then, what is the value of x ?

Explanation

To solve the problem, we need to find the tension in the string given the equation of the traveling wave and the linear density of the string.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Linear density (μ) = 4 kg/m
- Wave equation: \( y = 0.07 \cos(250t - 15x) \)

2. **Extract the Coefficients:**
- From the wave equation, we can identify the coefficients:
- Coefficient of \( t \) (ω) = 250
- Coefficient of \( x \) (k) = 15

3. **Calculate the Wave Velocity (v):**
- The wave velocity \( v \) can be calculated using the formula:
\[ v = \frac{\omega}{k} \]
- Substituting the values:
\[ v = \frac{250}{15} = \frac{50}{3} \text{ m/s} \]

4. **Use the Wave Velocity Formula:**
- The relationship between wave velocity, tension (T), and linear density (μ) is given by:
\[ v = \sqrt{\frac{T}{\mu}} \]
- Rearranging this formula to find tension \( T \):
\[ T = \mu v^2 \]

5. **Substitute the Values:**
- We already have \( \mu = 4 \text{ kg/m} \) and \( v = \frac{50}{3} \text{ m/s} \).
- Now calculate \( v^2 \):
\[ v^2 = \left(\frac{50}{3}\right)^2 = \frac{2500}{9} \text{ m}^2/\text{s}^2 \]
- Now substitute into the tension formula:
\[ T = 4 \cdot \frac{2500}{9} = \frac{10000}{9} \text{ N} \]

6. **Calculate the Final Value of Tension:**
- To find the numerical value:
\[ T \approx 1111.11 \text{ N} \]

### Conclusion:
The value of \( x \) (the tension in the string) is approximately \( 1111.11 \text{ N} \).