A wire is stretched between two posts. Another wire is stretched between two posts that are twice as far apart. The tension in the wires is the same, and they have the same mass. A transverse wave travels on the shorter wire with a speed of 240m/s. What would be the speed of the wave on the longer wire?

Options

A. 170m/s
B. 240m/s
C. 290m/s
D. 340m/s

Answer

To solve the problem, we need to find the speed of a transverse wave on a longer wire that is stretched between two posts that are twice as far apart as the shorter wire. We know the speed of the wave on the shorter wire and that the tension in both wires is the same, and they have the same mass.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Speed of wave on shorter wire (V1) = 240 m/s
- Length of shorter wire (L) = L
- Length of longer wire (L2) = 2L (twice the length of the shorter wire)
- Tension (T) in both wires = same
- Mass (m) of both wires = same

2. **Understand the Wave Speed Formula:**
The speed of a transverse wave on a wire is given by the formula:
\[ V = \sqrt{\frac{T}{\mu}} \]
where \(T\) is the tension in the wire and \(\mu\) is the mass per unit length of the wire.

3. **Calculate Mass per Unit Length for Both Wires:**
The mass per unit length (\(\mu\)) is defined as:
\[ \mu = \frac{m}{L} \]
For the shorter wire:
\[ \mu_1 = \frac{m}{L} \]
For the longer wire:
\[ \mu_2 = \frac{m}{2L} \quad (\text{since the length is doubled}) \]

4. **Express the Wave Speeds:**
For the shorter wire:
\[ V_1 = \sqrt{\frac{T}{\mu_1}} = \sqrt{\frac{T}{\frac{m}{L}}} = \sqrt{\frac{TL}{m}} \]
For the longer wire:
\[ V_2 = \sqrt{\frac{T}{\mu_2}} = \sqrt{\frac{T}{\frac{m}{2L}}} = \sqrt{\frac{2TL}{m}} \]

5. **Relate the Two Speeds:**
Now, we can relate the speeds \(V_1\) and \(V_2\):
\[ V_2 = \sqrt{2} \cdot V_1 \]

6. **Substitute the Known Value:**
Now substitute \(V_1 = 240 \, \text{m/s}\):
\[ V_2 = \sqrt{2} \cdot 240 \approx 1.414 \cdot 240 \approx 339.4 \, \text{m/s} \]

### Final Answer:
The speed of the wave on the longer wire is approximately **339.4 m/s**.