A copper rod of length 1.0 m is clamped at its middle point. Find the frequencies between 20 Hz and 20,000 Hz at which standing longitudinal waves can be set up in the rod. The speed of sound in copper is `3.8 km s^-1`.

Explanation

To find the frequencies at which standing longitudinal waves can be set up in a copper rod of length 1.0 m clamped at its middle point, we can follow these steps:

### Step 1: Understand the setup
The copper rod is clamped at its middle, which means it will vibrate in such a way that there are nodes at the clamped ends and antinodes at the free ends. The length of the rod is 1.0 m, and since it is clamped at the middle, each half of the rod (L/2) will vibrate.

### Step 2: Determine the wavelength
For a rod clamped at the middle, the fundamental mode of vibration will have a wavelength (λ) that is twice the length of each half of the rod. Therefore, we can express the relationship as:
\[ \frac{\lambda}{2} = \frac{L}{2} \]
This leads to:
\[ \lambda = 2L \]
Substituting the length of the rod (L = 1.0 m):
\[ \lambda = 2 \times 1.0 \, \text{m} = 2.0 \, \text{m} \]

### Step 3: Calculate the fundamental frequency
The speed of sound in copper is given as \( v = 3.8 \, \text{km/s} = 3800 \, \text{m/s} \). The frequency (f) can be calculated using the formula:
\[ f = \frac{v}{\lambda} \]
Substituting the values:
\[ f = \frac{3800 \, \text{m/s}}{2.0 \, \text{m}} = 1900 \, \text{Hz} \]

### Step 4: Determine higher harmonics
The frequencies of the higher harmonics can be expressed as:
\[ f_n = n \cdot f_0 \]
where \( f_0 \) is the fundamental frequency and \( n \) is a positive integer (1, 2, 3,...). Thus:
\[ f_n = n \cdot 1900 \, \text{Hz} \]

### Step 5: Find the range of n
We need to find the values of n such that the frequencies lie between 20 Hz and 20,000 Hz:
1. For \( n = 1 \):
\[ f_1 = 1 \cdot 1900 = 1900 \, \text{Hz} \]
2. For \( n = 2 \):
\[ f_2 = 2 \cdot 1900 = 3800 \, \text{Hz} \]
3. For \( n = 3 \):
\[ f_3 = 3 \cdot 1900 = 5700 \, \text{Hz} \]
4. For \( n = 4 \):
\[ f_4 = 4 \cdot 1900 = 7600 \, \text{Hz} \]
5. For \( n = 5 \):
\[ f_5 = 5 \cdot 1900 = 9500 \, \text{Hz} \]
6. For \( n = 6 \):
\[ f_6 = 6 \cdot 1900 = 11400 \, \text{Hz} \]
7. For \( n = 7 \):
\[ f_7 = 7 \cdot 1900 = 13300 \, \text{Hz} \]
8. For \( n = 8 \):
\[ f_8 = 8 \cdot 1900 = 15200 \, \text{Hz} \]
9. For \( n = 9 \):
\[ f_9 = 9 \cdot 1900 = 17100 \, \text{Hz} \]
10. For \( n = 10 \):
\[ f_{10} = 10 \cdot 1900 = 19000 \, \text{Hz} \]

### Step 6: List the possible frequencies
The possible frequencies between 20 Hz and 20,000 Hz are:
- 1900 Hz
- 3800 Hz
- 5700 Hz
- 7600 Hz
- 9500 Hz
- 11400 Hz
- 13300 Hz
- 15200 Hz
- 17100 Hz
- 19000 Hz

### Final Answer
The frequencies at which standing longitudinal waves can be set up in the copper rod are:
- 1900 Hz
- 3800 Hz
- 5700 Hz
- 7600 Hz
- 9500 Hz
- 11400 Hz
- 13300 Hz
- 15200 Hz
- 17100 Hz
- 19000 Hz