The equation of a transverse wave on a string is
`y =(2.0 mm) sin[(15 m^(-1)]x -(900 s^(-1))t].` The linear density is 4.17 g/m. (a) What is the wave speed? (b) What is the tension in the string?

Answer

To solve the problem, we will break it down into two parts: finding the wave speed and then calculating the tension in the string.

### Step-by-Step Solution

**Given:**
- Wave equation: \( y = (2.0 \, \text{mm}) \sin[(15 \, \text{m}^{-1})x - (900 \, \text{s}^{-1})t] \)
- Linear density (\( \mu \)): \( 4.17 \, \text{g/m} \)

**(a) Finding the Wave Speed:**

1. **Identify the angular frequency (\( \omega \)) and wave number (\( k \)) from the wave equation.**
- From the equation, we see that:
- \( \omega = 900 \, \text{s}^{-1} \)
- \( k = 15 \, \text{m}^{-1} \)

2. **Use the relationship between wave speed (\( v \)), angular frequency (\( \omega \)), and wave number (\( k \)).**
- The formula is:
\[ v = \frac{\omega}{k} \]

3. **Substitute the values of \( \omega \) and \( k \) into the formula.**
\[ v = \frac{900 \, \text{s}^{-1}}{15 \, \text{m}^{-1}} = 60 \, \text{m/s} \]

**(b) Finding the Tension in the String:**

1. **Convert the linear density from grams per meter to kilograms per meter.**
- Since \( 1 \, \text{g} = 0.001 \, \text{kg} \):
\[ \mu = 4.17 \, \text{g/m} = 0.00417 \, \text{kg/m} \]

2. **Use the relationship between wave speed (\( v \)), tension (\( T \)), and linear density (\( \mu \)).**
- The formula is:
\[ v = \sqrt{\frac{T}{\mu}} \]

3. **Square both sides to eliminate the square root.**
\[ v^2 = \frac{T}{\mu} \]

4. **Rearrange the formula to solve for tension (\( T \)).**
\[ T = v^2 \cdot \mu \]

5. **Substitute the values of \( v \) and \( \mu \) into the formula.**
\[ T = (60 \, \text{m/s})^2 \cdot 0.00417 \, \text{kg/m} = 3600 \cdot 0.00417 \approx 15.012 \, \text{N} \]

6. **Round the tension to two significant figures.**
\[ T \approx 15 \, \text{N} \]

### Final Answers:
- (a) Wave speed \( v = 60 \, \text{m/s} \)
- (b) Tension in the string \( T \approx 15 \, \text{N} \)

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