Two wires of equal length are stretched by the same force. One wire is heavy while the other is light which one will have less frequency

Options

A. heavier wire
B. lighter wire
C. both have same frequency
D. data insufficient

Answer

To solve the problem of determining which wire will have less frequency when two wires of equal length are stretched by the same force, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have two wires of equal length (L1 = L2).
- They are subjected to the same tensile force (F1 = F2).
- One wire is heavier (mass M1) and the other is lighter (mass M2), with M1 > M2.

2. **Frequency of Vibrating Strings**:
- The frequency of a vibrating wire is determined by its mass and tension. The formula for the fundamental frequency (f) of a wire is given by:
\[ f \propto \frac{1}{\sqrt{m}} \]
where \( m \) is the mass per unit length (linear density) of the wire.

3. **Relating Mass and Frequency**:
- Since both wires are of equal length and are stretched by the same force, we can say:
\[ f_1 \propto \frac{1}{\sqrt{M_1}} \quad \text{(for the heavier wire)} \]
\[ f_2 \propto \frac{1}{\sqrt{M_2}} \quad \text{(for the lighter wire)} \]

4. **Comparing Frequencies**:
- Given that M1 > M2, we can analyze the relationship:
- Since \( M_1 \) (mass of the heavier wire) is greater than \( M_2 \) (mass of the lighter wire), it follows that:
\[ \sqrt{M_1} > \sqrt{M_2} \]
- Therefore, we can conclude:
\[ f_1 < f_2 \]
- This means that the frequency of the heavier wire (f1) is less than the frequency of the lighter wire (f2).

5. **Final Conclusion**:
- Since the heavier wire has a lower frequency compared to the lighter wire, the answer to the question is that the heavier wire will have less frequency.

### Answer:
The heavier wire will have less frequency.

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