A tube of a certain diameter and of length `48cm` is open at both ends. Its fundamental frequency is found to be `320 Hz`. The velocity of sound in air is `320 m//sec`. Estimate the diameter of the tube.
One end of the tube is now closed. Calculate the lowest frequrncy of resonance for the tube.

Explanation

To solve the problem step by step, we will first find the diameter of the tube when it is open at both ends, and then we will calculate the lowest frequency of resonance when one end of the tube is closed.

### Step 1: Understand the relationship between frequency, velocity, and length for a tube open at both ends.

For a tube open at both ends, the fundamental frequency (first harmonic) is given by the formula:

\[ f = \frac{v}{2L} \]

where:
- \( f \) is the fundamental frequency,
- \( v \) is the velocity of sound in air,
- \( L \) is the length of the tube.

### Step 2: Incorporate the end correction.

Since the tube has an end correction, we need to modify the length \( L \) to account for this. The end correction \( E \) is given by:

\[ E = 0.6D \]

Thus, the effective length of the tube becomes:

\[ L_{\text{effective}} = L + E = L + 0.6D \]

### Step 3: Set up the equation with the given values.

We know:
- \( f = 320 \, \text{Hz} \)
- \( v = 320 \, \text{m/s} \)
- \( L = 48 \, \text{cm} = 0.48 \, \text{m} \)

Substituting these values into the frequency equation gives:

\[ 320 = \frac{320}{2(0.48 + 0.6D)} \]

### Step 4: Solve for \( D \).

Rearranging the equation:

\[ 320(2(0.48 + 0.6D)) = 320 \]

This simplifies to:

\[ 2(0.48 + 0.6D) = 1 \]

Dividing both sides by 2:

\[ 0.48 + 0.6D = 0.5 \]

Now, isolate \( D \):

\[ 0.6D = 0.5 - 0.48 = 0.02 \]

\[ D = \frac{0.02}{0.6} = 0.0333 \, \text{m} = 3.33 \, \text{cm} \]

### Step 5: Calculate the lowest frequency of resonance when one end is closed.

For a tube closed at one end, the fundamental frequency is given by:

\[ f = \frac{v}{4L_{\text{effective}}} \]

The end correction for a closed end is \( E = 0.3D \), so:

\[ L_{\text{effective}} = L + 0.3D \]

Substituting the values we have:

\[ L_{\text{effective}} = 0.48 + 0.3(0.0333) = 0.48 + 0.01 = 0.49 \, \text{m} \]

Now, substituting into the frequency formula:

\[ f = \frac{320}{4(0.49)} = \frac{320}{1.96} \approx 163.27 \, \text{Hz} \]

### Final Answers:

1. The diameter of the tube is \( D = 3.33 \, \text{cm} \).
2. The lowest frequency of resonance when one end is closed is approximately \( f \approx 163 \, \text{Hz} \).