The mass of a string is `5.0 xx 10^(-3) kg,` and it is stretched so that the tension in it is 180 N. A transverse wave traveling on this string has a frequency of 260 Hz and a wavelength: of 0.60 m. What is the length of the string?

Options

A. 1.1m
B. 0.68m
C. 0.47m
D. 0.22m

Answer

To solve the problem, we will use the relationship between the speed of a wave on a string, its tension, and its mass per unit length. We will also use the wave equation that relates frequency, wavelength, and speed.

### Step-by-Step Solution:

1. **Identify the given values:**
- Mass of the string, \( m = 5.0 \times 10^{-3} \, \text{kg} \)
- Tension in the string, \( T = 180 \, \text{N} \)
- Frequency of the wave, \( f = 260 \, \text{Hz} \)
- Wavelength of the wave, \( \lambda = 0.60 \, \text{m} \)

2. **Calculate the mass per unit length (\( \mu \)):**
\[ \mu = \frac{m}{L} \]
where \( L \) is the length of the string. We will find \( L \) later.

3. **Use the wave speed formula:**
The speed of the wave \( v \) can be calculated using the formula:
\[ v = f \cdot \lambda \]
Substituting the values:
\[ v = 260 \, \text{Hz} \times 0.60 \, \text{m} = 156 \, \text{m/s} \]

4. **Relate wave speed to tension and mass per unit length:**
The speed of the wave can also be expressed as:
\[ v = \sqrt{\frac{T}{\mu}} \]
Rearranging this gives:
\[ \mu = \frac{T}{v^2} \]

5. **Substituting the values:**
First, we calculate \( v^2 \):
\[ v^2 = (156 \, \text{m/s})^2 = 24336 \, \text{m}^2/\text{s}^2 \]
Now substituting \( T \) and \( v^2 \):
\[ \mu = \frac{180 \, \text{N}}{24336 \, \text{m}^2/\text{s}^2} \approx 0.00739 \, \text{kg/m} \]

6. **Using the mass per unit length to find the length of the string:**
From the definition of \( \mu \):
\[ \mu = \frac{m}{L} \implies L = \frac{m}{\mu} \]
Substituting the values:
\[ L = \frac{5.0 \times 10^{-3} \, \text{kg}}{0.00739 \, \text{kg/m}} \approx 0.678 \, \text{m} \]

7. **Final Result:**
The length of the string is approximately \( 0.68 \, \text{m} \).