A stretched rope having linear mass density `5xx10^(-2)kg//m` is under a tension of `80 N`. the power that has to be supoplied to the rope to generate harmonic waves at a frequency of `60 Hz` and an amplitude of `(2sqrt2)/(15pi)m` is

Options

A. 215 W
B. 251 w
C. 512 w
D. 521 w

Explanation

To solve the problem, we need to calculate the power that must be supplied to a stretched rope to generate harmonic waves. We will use the formula for power in a wave on a string, which is given by:

\[ P = \frac{1}{2} \mu \omega^2 A^2 v \]

where:
- \( P \) is the power,
- \( \mu \) is the linear mass density,
- \( \omega \) is the angular frequency,
- \( A \) is the amplitude of the wave,
- \( v \) is the wave velocity.

### Step 1: Identify the given values
- Linear mass density \( \mu = 5 \times 10^{-2} \, \text{kg/m} \)
- Tension \( T = 80 \, \text{N} \)
- Frequency \( f = 60 \, \text{Hz} \)
- Amplitude \( A = \frac{2\sqrt{2}}{15\pi} \, \text{m} \)

### Step 2: Calculate the wave velocity \( v \)
The wave velocity \( v \) on a string is given by:

\[ v = \sqrt{\frac{T}{\mu}} \]

Substituting the values:

\[ v = \sqrt{\frac{80 \, \text{N}}{5 \times 10^{-2} \, \text{kg/m}}} \]

Calculating this gives:

\[ v = \sqrt{\frac{80}{0.05}} = \sqrt{1600} = 40 \, \text{m/s} \]

### Step 3: Calculate the angular frequency \( \omega \)
The angular frequency \( \omega \) is given by:

\[ \omega = 2\pi f \]

Substituting the frequency:

\[ \omega = 2\pi \times 60 \]

Calculating this gives:

\[ \omega = 120\pi \, \text{rad/s} \]

### Step 4: Calculate the power \( P \)
Now we can substitute all the values into the power formula:

\[ P = \frac{1}{2} \mu \omega^2 A^2 v \]

Substituting the values:

\[ P = \frac{1}{2} \times (5 \times 10^{-2}) \times (120\pi)^2 \times \left(\frac{2\sqrt{2}}{15\pi}\right)^2 \times 40 \]

Calculating \( \omega^2 \):

\[ \omega^2 = (120\pi)^2 = 14400\pi^2 \]

Calculating \( A^2 \):

\[ A^2 = \left(\frac{2\sqrt{2}}{15\pi}\right)^2 = \frac{8}{225\pi^2} \]

Now substituting these into the power equation:

\[ P = \frac{1}{2} \times (5 \times 10^{-2}) \times (14400\pi^2) \times \left(\frac{8}{225\pi^2}\right) \times 40 \]

Simplifying this:

\[ P = \frac{1}{2} \times (5 \times 10^{-2}) \times 14400 \times \frac{8 \times 40}{225} \]

Calculating the constants:

\[ P = \frac{1}{2} \times (5 \times 10^{-2}) \times 14400 \times \frac{320}{225} \]

Calculating \( \frac{320}{225} \):

\[ \frac{320}{225} \approx 1.4222 \]

Now substituting back:

\[ P \approx \frac{1}{2} \times (5 \times 10^{-2}) \times 14400 \times 1.4222 \]

Calculating this step-by-step gives:

\[ P \approx 512 \, \text{W} \]

### Final Answer
The power that has to be supplied to the rope is approximately **512 Watts**.