Sinusoidal waves `5.00 cm ` in amplitude are to be transmitted along a string having a linear mass density equal to `4.00xx10^-2kg//m`. If the source can deliver a maximum power of `90W` and the string is under a tension of `100N`, then the highest frequency at which the source can operate is (take `pi^2=10`)

Options

A. 45.3 Hz
B. 50 Hz
C. 30 Hz
D. 62.3 Hz

Explanation

To find the highest frequency at which the source can operate, we can use the relationship between power, amplitude, linear mass density, tension, and frequency for a sinusoidal wave on a string. Here’s a step-by-step solution:

### Step 1: Write down the formula for power transmitted by a wave on a string.
The power \( P \) transmitted by a sinusoidal wave on a string is given by the formula:
\[ P = \frac{1}{2} \mu A^2 \omega^2 v \]
where:
- \( P \) = power (in watts)
- \( \mu \) = linear mass density (in kg/m)
- \( A \) = amplitude (in meters)
- \( \omega \) = angular frequency (in rad/s)
- \( v \) = wave speed (in m/s)

### Step 2: Write the expression for wave speed \( v \).
The wave speed \( v \) on a string under tension \( T \) is given by:
\[ v = \sqrt{\frac{T}{\mu}} \]
where \( T \) is the tension in the string.

### Step 3: Substitute \( v \) into the power equation.
Substituting the expression for \( v \) into the power equation, we get:
\[ P = \frac{1}{2} \mu A^2 \omega^2 \sqrt{\frac{T}{\mu}} \]
This simplifies to:
\[ P = \frac{1}{2} A^2 \omega^2 \sqrt{\mu T} \]

### Step 4: Solve for angular frequency \( \omega \).
Rearranging the equation for \( \omega^2 \):
\[ \omega^2 = \frac{2P}{A^2 \sqrt{\mu T}} \]
Taking the square root gives:
\[ \omega = \sqrt{\frac{2P}{A^2 \sqrt{\mu T}}} \]

### Step 5: Relate angular frequency to frequency.
We know that:
\[ \omega = 2\pi f \]
Thus,
\[ f = \frac{\omega}{2\pi} = \frac{1}{2\pi} \sqrt{\frac{2P}{A^2 \sqrt{\mu T}}} \]

### Step 6: Substitute the values into the frequency formula.
Given:
- \( P = 90 \, \text{W} \)
- \( A = 5.00 \, \text{cm} = 0.05 \, \text{m} \)
- \( \mu = 4.00 \times 10^{-2} \, \text{kg/m} \)
- \( T = 100 \, \text{N} \)

Substituting these values:
\[ f = \frac{1}{2\pi} \sqrt{\frac{2 \times 90}{(0.05)^2 \sqrt{(4.00 \times 10^{-2})(100)}}} \]

### Step 7: Calculate the values.
First, calculate \( \sqrt{\mu T} \):
\[ \sqrt{\mu T} = \sqrt{(4.00 \times 10^{-2})(100)} = \sqrt{4} = 2 \]
Now substitute this into the frequency equation:
\[ f = \frac{1}{2\pi} \sqrt{\frac{180}{(0.05)^2 \cdot 2}} \]
Calculating \( (0.05)^2 = 0.0025 \):
\[ f = \frac{1}{2\pi} \sqrt{\frac{180}{0.0025 \cdot 2}} = \frac{1}{2\pi} \sqrt{\frac{180}{0.005}} = \frac{1}{2\pi} \sqrt{36000} \]
Calculating \( \sqrt{36000} = 189.736 \):
\[ f = \frac{189.736}{2\pi} \approx \frac{189.736}{6.2832} \approx 30.2 \, \text{Hz} \]

### Step 8: Final result.
Thus, the highest frequency at which the source can operate is approximately:
\[ f \approx 30 \, \text{Hz} \]