A copper wire, whose cross-sectional area is `1.1 xx 10^(-6) m^2`, has a linear density of `70 xx 10^(-3)` kg/m and is strung between two walls. At the ambient temperature, a transverse wave travels with a speed of 46 m/s on this wire. The coefficient of linear expansion for copper is `17 xx 10^(-6)(""^(@)C),` and Young's modulus for copper is `1.1 xx 10^(11) N//m^2.` What will be the speed of the wave when the temperature is lowered by `14^@C?` Ignore any change in the linear density caused by the change in temperature.

Options

A. 84m/s
B. 79m/s
C. 61m/s
D. 55m/s

Answer

To find the speed of the wave when the temperature is lowered by \(14^\circ C\), we can follow these steps:

### Step 1: Identify Given Values
- Cross-sectional area \(A = 1.1 \times 10^{-6} \, m^2\)
- Linear density \(\lambda = 70 \times 10^{-3} \, kg/m\)
- Initial wave speed \(v_1 = 46 \, m/s\)
- Coefficient of linear expansion for copper \(\alpha = 17 \times 10^{-6} \, /{}^\circ C\)
- Young's modulus for copper \(Y = 1.1 \times 10^{11} \, N/m^2\)
- Change in temperature \(\Delta T = -14 \, ^\circ C\)

### Step 2: Calculate Initial Tension in the Wire
The speed of a wave on a string is given by the formula:
\[ v = \sqrt{\frac{F}{\lambda}} \]
where \(F\) is the tension in the wire and \(\lambda\) is the linear density.

Rearranging this formula to find the tension \(F\):
\[ F = v^2 \cdot \lambda \]
Substituting the known values:
\[ F_1 = (46 \, m/s)^2 \cdot (70 \times 10^{-3} \, kg/m) \]
Calculating \(F_1\):
\[ F_1 = 2116 \cdot 0.070 = 148.12 \, N \]

### Step 3: Calculate Change in Length Due to Temperature Change
The change in length due to temperature change can be expressed as:
\[ \Delta L = \alpha \cdot L_0 \cdot \Delta T \]
The new length \(L\) will be:
\[ L = L_0 (1 + \alpha \Delta T) \]
The strain is given by:
\[ \text{Strain} = \frac{\Delta L}{L_0} = \alpha \Delta T \]

### Step 4: Calculate the New Tension
Using Young's modulus, we can find the new force due to the change in temperature:
\[ Y = \frac{F \cdot L_0}{A \cdot \Delta L} \]
Rearranging gives:
\[ F = Y \cdot A \cdot \alpha \cdot \Delta T \]
Substituting the values:
\[ F = (1.1 \times 10^{11} \, N/m^2) \cdot (1.1 \times 10^{-6} \, m^2) \cdot (17 \times 10^{-6} \, /{}^\circ C) \cdot (-14 \, ^\circ C) \]
Calculating \(F\):
\[ F = 1.1 \times 10^{11} \cdot 1.1 \times 10^{-6} \cdot 17 \times 10^{-6} \cdot (-14) \]
\[ F \approx -28.7 \, N \]
(Note: The negative sign indicates a decrease in tension due to cooling.)

### Step 5: Calculate Total Tension
The new tension \(F_2\) in the wire will be:
\[ F_2 = F_1 + F = 148.12 \, N + (-28.7 \, N) = 119.42 \, N \]

### Step 6: Calculate New Wave Speed
Using the new tension to find the new wave speed \(v_2\):
\[ v_2 = \sqrt{\frac{F_2}{\lambda}} \]
Substituting the values:
\[ v_2 = \sqrt{\frac{119.42 \, N}{70 \times 10^{-3} \, kg/m}} \]
Calculating \(v_2\):
\[ v_2 = \sqrt{\frac{119.42}{0.070}} \approx \sqrt{1706} \approx 50.25 \, m/s \]

### Final Answer
The speed of the wave when the temperature is lowered by \(14^\circ C\) is approximately \(50.25 \, m/s\).

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