To solve the problem, we need to determine the natural frequencies of a copper rod clamped at its midpoint and find how many of these frequencies fall within the range of 20 kHz to 50 kHz.
### Step-by-Step Solution:
1. **Identify the Length of the Rod**:
The length of the copper rod is given as \( l = 50 \, \text{cm} = 0.5 \, \text{m} \).
2. **Understand the Clamping Condition**:
Since the rod is clamped at its midpoint, it will have nodes at the midpoint and oscillate in a longitudinal manner. The fundamental frequency and harmonics will be determined based on this condition.
3. **Determine the Wavelength**:
For a rod clamped at its midpoint, the wavelengths of the oscillations can be expressed as:
\[
\lambda_n = \frac{2l}{n}
\]
where \( n \) is an odd integer (1, 3, 5, ...).
4. **Calculate the Velocity of Sound in Copper**:
The speed of sound in copper is given as \( v = 3.8 \, \text{km/s} = 3800 \, \text{m/s} \).
5. **Relate Frequency and Wavelength**:
The frequency \( f \) is related to the wavelength \( \lambda \) and the speed \( v \) by the equation:
\[
f = \frac{v}{\lambda}
\]
Substituting the expression for \( \lambda_n \):
\[
f_n = \frac{v}{\frac{2l}{n}} = \frac{nv}{2l}
\]
Since \( n \) must be odd, we can express this as:
\[
f_n = \frac{(2n + 1)v}{2l}
\]
6. **Substituting Values**:
Substituting \( v = 3800 \, \text{m/s} \) and \( l = 0.5 \, \text{m} \):
\[
f_n = \frac{(2n + 1) \cdot 3800}{2 \cdot 0.5} = (2n + 1) \cdot 3800 \, \text{Hz}
\]
\[
f_n = (2n + 1) \cdot 3800 \, \text{Hz}
\]
7. **Finding Frequencies**:
Now we calculate the frequencies for odd values of \( n \):
- For \( n = 0 \): \( f_0 = 1 \cdot 3800 = 3800 \, \text{Hz} = 3.8 \, \text{kHz} \)
- For \( n = 1 \): \( f_1 = 3 \cdot 3800 = 11400 \, \text{Hz} = 11.4 \, \text{kHz} \)
- For \( n = 2 \): \( f_2 = 5 \cdot 3800 = 19000 \, \text{Hz} = 19.0 \, \text{kHz} \)
- For \( n = 3 \): \( f_3 = 7 \cdot 3800 = 26600 \, \text{Hz} = 26.6 \, \text{kHz} \)
- For \( n = 4 \): \( f_4 = 9 \cdot 3800 = 34200 \, \text{Hz} = 34.2 \, \text{kHz} \)
- For \( n = 5 \): \( f_5 = 11 \cdot 3800 = 41800 \, \text{Hz} = 41.8 \, \text{kHz} \)
- For \( n = 6 \): \( f_6 = 13 \cdot 3800 = 49400 \, \text{Hz} = 49.4 \, \text{kHz} \)
- For \( n = 7 \): \( f_7 = 15 \cdot 3800 = 57000 \, \text{Hz} = 57.0 \, \text{kHz} \) (exceeds 50 kHz)
8. **Count Frequencies in the Range**:
The frequencies that fall within the range of 20 kHz to 50 kHz are:
- 26.6 kHz
- 34.2 kHz
- 41.8 kHz
- 49.4 kHz
Thus, there are **4 natural longitudinal oscillations** of the rod in the frequency range from 20 kHz to 50 kHz.
### Final Answer:
The number of natural longitudinal oscillations of the rod in the frequency range from 20 kHz to 50 kHz is **4**, and the corresponding frequencies are **26.6 kHz, 34.2 kHz, 41.8 kHz, and 49.4 kHz**.
