A string 2.0 m long and fixed at its ends is driven by a 240 Hz vibrtor. The string vibrates in its third harmonic mode. The speed of the wave and its fundamental frequency is :

Options

A. 320 m/s, 80 Hz
B. 320 m/s, 120 Hz
C. 180 m/s, 80 Hz
D. 180 m/s, 120 Hz

Answer

To solve the problem, we need to find the speed of the wave on the string and its fundamental frequency. Let's break it down step by step.

### Step-by-Step Solution:

1. **Identify the Length of the String and the Harmonic Mode:**
- The string is 2.0 m long and is vibrating in its third harmonic mode.
- In the third harmonic, there are 3 half-wavelengths fitting into the length of the string.

2. **Determine the Wavelength:**
- The relationship between the length of the string (L) and the wavelength (λ) in the nth harmonic is given by:
\[ L = \frac{n \cdot \lambda}{2} \]
- For the third harmonic (n = 3):
\[ 2.0 = \frac{3 \cdot \lambda}{2} \]
- Rearranging gives:
\[ \lambda = \frac{2 \cdot 2.0}{3} = \frac{4.0}{3} \text{ m} \]

3. **Calculate the Frequency:**
- The string is driven by a 240 Hz vibrator, which corresponds to the frequency of the third harmonic:
\[ f_3 = 240 \text{ Hz} \]
- The fundamental frequency (f₀) is related to the third harmonic by:
\[ f_3 = 3 \cdot f_0 \]
- Therefore:
\[ f_0 = \frac{f_3}{3} = \frac{240}{3} = 80 \text{ Hz} \]

4. **Calculate the Wave Speed:**
- The speed of the wave (v) on the string can be calculated using the formula:
\[ v = f \cdot \lambda \]
- Using the frequency of the third harmonic (240 Hz) and the wavelength we found:
\[ v = 240 \cdot \frac{4.0}{3} = 320 \text{ m/s} \]

5. **Final Results:**
- The fundamental frequency (f₀) is 80 Hz.
- The speed of the wave (v) is 320 m/s.

### Summary of Results:
- **Fundamental Frequency (f₀):** 80 Hz
- **Wave Speed (v):** 320 m/s