A string of length 2m is fixed at two ends. It is in resonance with a tuning fork of frequency 240 Hz in its third harmonic. Then speed of wave sound in string and its fundamental frequency is:

Options

A. `240m//s,80Hz`
B. `320m//s,80Hz`
C. `1640m//s,80Hz`
D. `120m//s,40Hz`

Answer

To solve the problem step by step, we will follow the reasoning outlined in the video transcript.

### Step 1: Understand the problem
We have a string of length \( L = 2 \, \text{m} \) that is fixed at both ends. It is vibrating in its third harmonic mode when it resonates with a tuning fork of frequency \( f = 240 \, \text{Hz} \).

### Step 2: Relate the length of the string to the wavelength
In the third harmonic, the string forms three segments (or loops). The relationship between the length of the string and the wavelength \( \lambda \) in the third harmonic can be expressed as:
\[ L = \frac{3}{2} \lambda \]
Substituting the length of the string:
\[ 2 = \frac{3}{2} \lambda \]

### Step 3: Solve for the wavelength
To find the wavelength \( \lambda \), we can rearrange the equation:
\[ \lambda = \frac{2 \times 2}{3} = \frac{4}{3} \, \text{m} \]

### Step 4: Calculate the speed of the wave in the string
The speed of the wave \( v \) in the string can be calculated using the formula:
\[ v = f \cdot \lambda \]
Substituting the values we have:
\[ v = 240 \, \text{Hz} \times \frac{4}{3} \, \text{m} = 320 \, \text{m/s} \]

### Step 5: Find the fundamental frequency
The fundamental frequency \( f_1 \) is related to the frequency of the third harmonic \( f_3 \) by the formula:
\[ f_n = n \cdot f_1 \]
For the third harmonic (\( n = 3 \)):
\[ f_3 = 3 \cdot f_1 \]
Thus, we can find the fundamental frequency:
\[ f_1 = \frac{f_3}{3} = \frac{240 \, \text{Hz}}{3} = 80 \, \text{Hz} \]

### Final Answers
- Speed of the wave in the string: \( 320 \, \text{m/s} \)
- Fundamental frequency: \( 80 \, \text{Hz} \)