A uniform tube of length `60 cm` stands vertically with its lower end dipping into water . First two air column lengths above water are `15 cm and 45 cm`, when the tube responds to a vibrating fork of frequency `500 H_(Z)` . Find the lowest frequency to the tube will respond when it is open at both ends.

Explanation

To solve the problem step by step, we will follow these procedures:

### Step 1: Understand the Problem
We have a uniform tube of length 60 cm, with its lower end dipping into water. The tube behaves as a closed pipe at one end (the end in water) and we are given the first two resonant lengths of the air column above the water: 15 cm and 45 cm. The frequency of the vibrating fork is 500 Hz. We need to find the lowest frequency when the tube is open at both ends.

### Step 2: Use the Formula for Closed Pipes
For a closed pipe (one end closed), the fundamental frequency (first harmonic) is given by the formula:
\[ f = \frac{V}{4L} \]
where:
- \( f \) is the frequency,
- \( V \) is the speed of sound in air (approximately 300 m/s),
- \( L \) is the length of the air column.

### Step 3: Calculate the Speed of Sound
Using the first air column length (15 cm):
1. Convert 15 cm to meters: \( L = 0.15 \, \text{m} \).
2. Substitute into the formula:
\[ 500 = \frac{V}{4 \times 0.15} \]
3. Rearranging gives:
\[ V = 500 \times 4 \times 0.15 = 300 \, \text{m/s} \]

### Step 4: Verify with Second Air Column Length
Now, use the second air column length (45 cm):
1. Convert 45 cm to meters: \( L = 0.45 \, \text{m} \).
2. Substitute into the formula:
\[ f = \frac{V}{4L} = \frac{300}{4 \times 0.45} \]
3. Calculate:
\[ f = \frac{300}{1.8} = 166.67 \, \text{Hz} \]
This value is not needed for the final answer but confirms our speed of sound is consistent.

### Step 5: Calculate Frequency for Open Pipe
For an open pipe (open at both ends), the fundamental frequency is given by:
\[ f = \frac{V}{2L} \]
Here, the length \( L \) is the total length of the tube, which is 60 cm or 0.60 m.

### Step 6: Substitute Values
1. Substitute into the formula:
\[ f = \frac{300}{2 \times 0.60} \]
2. Calculate:
\[ f = \frac{300}{1.2} = 250 \, \text{Hz} \]

### Final Answer
The lowest frequency to which the tube will respond when it is open at both ends is:
\[ \text{250 Hz} \]

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