A transverse wave, y = −0.5sin(x + 45t) is produced in a wire of mass per unit length 0.12 g/cm, where x is in meters and t is in seconds. Then the expected value of tension in the wire is ______.

Options

A. 16.7 N
B. 24.3 N
C. 29.2 N
D. 26.7 N

Answer

To solve the problem, we need to determine the tension in the wire based on the given wave equation and the mass per unit length of the wire. Here’s the step-by-step solution:

### Step 1: Identify the wave parameters
The wave equation given is:
\[ y = -0.5 \sin(x + 45t) \]

We can rewrite this as:
\[ y = 0.5 \sin(-x - 45t) \]
Using the property of sine, we can express it as:
\[ y = 0.5 \sin(-1 \cdot x - 45t) \]

From this equation, we can identify:
- Amplitude \( A = 0.5 \) m
- Wave number \( k = -1 \) (taking the magnitude, \( |k| = 1 \))
- Angular frequency \( \omega = 45 \) rad/s

### Step 2: Calculate the wave velocity
The wave velocity \( v \) can be calculated using the formula:
\[ v = \frac{\omega}{k} \]
Substituting the values:
\[ v = \frac{45}{1} = 45 \text{ m/s} \]

### Step 3: Convert mass per unit length
The mass per unit length \( \mu \) is given as \( 0.12 \) g/cm. We need to convert this to kg/m:
\[ \mu = 0.12 \text{ g/cm} = 0.12 \times \frac{1 \text{ kg}}{1000 \text{ g}} \times \frac{100 \text{ cm}}{1 \text{ m}} = 0.012 \text{ kg/m} \]

### Step 4: Use the wave velocity to find tension
The relationship between wave velocity, tension, and mass per unit length is given by:
\[ v = \sqrt{\frac{T}{\mu}} \]
Squaring both sides gives:
\[ v^2 = \frac{T}{\mu} \]
Rearranging for tension \( T \):
\[ T = \mu v^2 \]

### Step 5: Substitute values to find tension
Substituting the values of \( \mu \) and \( v \):
\[ T = 0.012 \text{ kg/m} \times (45 \text{ m/s})^2 \]
Calculating \( (45)^2 \):
\[ (45)^2 = 2025 \]
Now substituting this back:
\[ T = 0.012 \times 2025 = 24.3 \text{ N} \]

### Final Answer
The expected value of tension in the wire is:
\[ \boxed{24.3 \text{ N}} \]