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 frequency of resonance for the tube.

Explanation

To solve the problem step by step, we will first estimate 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: Understanding the Fundamental Frequency of an Open Tube
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: Including End Correction
Since the tube has a finite diameter, we need to include an end correction. The end correction for a tube open at both ends is approximately \( 0.6D \), where \( D \) is the diameter of the tube. Therefore, the effective length \( L' \) of the tube becomes:
\[ L' = L + 0.6D \]
Thus, the modified formula for the fundamental frequency becomes:
\[ f = \frac{v}{2(L + 0.6D)} \]

### Step 3: Substituting Known Values
Given:
- \( f = 320 \, \text{Hz} \)
- \( v = 320 \, \text{m/s} \)
- \( L = 48 \, \text{cm} = 0.48 \, \text{m} \)

Substituting these values into the frequency formula:
\[ 320 = \frac{320}{2(0.48 + 0.6D)} \]

### Step 4: Simplifying the Equation
Cross-multiplying gives:
\[ 320 \cdot 2(0.48 + 0.6D) = 320 \]
This simplifies to:
\[ 640(0.48 + 0.6D) = 320 \]
Dividing both sides by 640:
\[ 0.48 + 0.6D = \frac{320}{640} = 0.5 \]

### Step 5: Solving for Diameter \( D \)
Rearranging the equation:
\[ 0.6D = 0.5 - 0.48 \]
\[ 0.6D = 0.02 \]
\[ D = \frac{0.02}{0.6} = 0.0333 \, \text{m} \approx 0.033 \, \text{m} \]

### Step 6: Calculating the Lowest Frequency of Resonance for Closed Tube
For a tube closed at one end, the fundamental frequency is given by:
\[ f' = \frac{v}{4L'} \]
where the end correction for a closed end is approximately \( 0.3D \). Thus, the effective length \( L' \) becomes:
\[ L' = L + 0.3D \]

### Step 7: Substituting Values for Closed Tube
Substituting the values:
\[ L' = 0.48 + 0.3 \times 0.0333 \]
Calculating \( 0.3 \times 0.0333 \):
\[ 0.3 \times 0.0333 \approx 0.01 \]
Thus,
\[ L' \approx 0.48 + 0.01 = 0.49 \, \text{m} \]

### Step 8: Calculating the New Frequency
Substituting into the frequency formula for the closed tube:
\[ f' = \frac{320}{4 \times 0.49} \]
Calculating:
\[ f' = \frac{320}{1.96} \approx 163.27 \, \text{Hz} \approx 163 \, \text{Hz} \]

### Final Answers
1. The diameter of the tube \( D \approx 0.033 \, \text{m} \) or \( 3.3 \, \text{cm} \).
2. The lowest frequency of resonance for the tube when one end is closed is approximately \( 163 \, \text{Hz} \).