A long glass tube is held vertically in water . A tuning fork is struck and held over the tube . Strong resonances are observed at two successive lengths `0.50 m and 0.84 m` above the surface of water . If the velocity of sound is `340 m//s`, then the frequency of the tuning fork is

Options

A. 128 Hz
B. 256 Hz
C. 384 Hz
D. 500 Hz

Answer

To find the frequency of the tuning fork, we can follow these steps:

### Step 1: Identify the given data
- Velocity of sound (v) = 340 m/s
- Lengths of resonance (L1 and L2):
- L1 = 0.50 m
- L2 = 0.84 m

### Step 2: Calculate the difference in lengths
The difference in lengths (ΔL) between the two successive resonance points can be calculated as:
\[ \Delta L = L2 - L1 \]
\[ \Delta L = 0.84 \, \text{m} - 0.50 \, \text{m} = 0.34 \, \text{m} \]

### Step 3: Relate the difference in lengths to the wavelength
In a tube closed at one end, the difference in the lengths of the resonating air column corresponds to half the wavelength (λ/2):
\[ \frac{\lambda}{2} = \Delta L \]
Thus, we can express the wavelength as:
\[ \lambda = 2 \times \Delta L \]
Substituting the value of ΔL:
\[ \lambda = 2 \times 0.34 \, \text{m} = 0.68 \, \text{m} \]

### Step 4: Calculate the frequency using the wavelength
The frequency (f) of the tuning fork can be calculated using the formula:
\[ f = \frac{v}{\lambda} \]
Substituting the values of v and λ:
\[ f = \frac{340 \, \text{m/s}}{0.68 \, \text{m}} \]
Calculating the frequency:
\[ f = 500 \, \text{Hz} \]

### Conclusion
The frequency of the tuning fork is **500 Hz**.

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