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

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the relationship between power, amplitude, frequency, and tension in a wave on a string.

The power \( P \) transmitted by a sinusoidal wave on a string can be expressed as:
\[ P = \frac{1}{2} \mu A^2 \omega^2 v \]
where:
- \( P \) = power (in watts)
- \( \mu \) = linear mass density (in kg/m)
- \( A \) = amplitude (in meters)
- \( \omega \) = angular frequency (in radians/second)
- \( v \) = wave speed (in m/s)

### Step 2: Find the 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.

Now, substituting \( v \) into the power equation, we have:
\[ P = \frac{1}{2} \mu A^2 \omega^2 \sqrt{\frac{T}{\mu}} \]
This simplifies to:
\[ P = \frac{1}{2} A^2 \omega^2 \sqrt{\mu T} \]

### Step 4: Solve for \( \omega \).

Rearranging the equation to solve for \( \omega \):
\[ \omega^2 = \frac{2P}{A^2 \sqrt{\mu T}} \]
Taking the square root gives:
\[ \omega = \sqrt{\frac{2P}{A^2 \sqrt{\mu T}}} \]

### Step 5: Find the frequency \( f \).

The frequency \( f \) is related to angular frequency \( \omega \) by:
\[ f = \frac{\omega}{2\pi} \]
Substituting the expression for \( \omega \):
\[ f = \frac{1}{2\pi} \sqrt{\frac{2P}{A^2 \sqrt{\mu T}}} \]

### Step 6: Substitute the values.

Now we will substitute the given values into the equation:
- \( P = 90 \, \text{W} \)
- \( A = 5.00 \, \text{cm} = 0.05 \, \text{m} \)
- \( \mu = 4.00 \times 10^{-2} \, \text{kg/m} \)
- \( T = 100 \, \text{N} \)

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

### Step 7: Calculate \( \sqrt{4.00 \times 10^{-2} \times 100} \).

Calculating the term inside the square root:
\[ \sqrt{4.00 \times 10^{-2} \times 100} = \sqrt{4.00} = 2 \]

### Step 8: Substitute back into the frequency equation.

Now substituting this back:
\[ f = \frac{1}{2\pi} \sqrt{\frac{180}{(0.05)^2 \times 2}} \]
Calculating \( (0.05)^2 = 0.0025 \):
\[ f = \frac{1}{2\pi} \sqrt{\frac{180}{0.0025 \times 2}} = \frac{1}{2\pi} \sqrt{\frac{180}{0.005}} = \frac{1}{2\pi} \sqrt{36000} \]

### Step 9: Calculate \( \sqrt{36000} \).

Calculating \( \sqrt{36000} \):
\[ \sqrt{36000} = 60 \sqrt{10} \]

### Step 10: Final calculation for frequency.

Now substituting this back:
\[ f = \frac{1}{2\pi} \times 60 \sqrt{10} \]
Using \( \pi^2 = 10 \):
\[ f = \frac{60 \sqrt{10}}{2 \times 3.16} = \frac{60 \sqrt{10}}{6.28} \approx 30 \, \text{Hz} \]

### Final Answer:
The highest frequency at which the source can operate is approximately \( 30 \, \text{Hz} \).