A taut string for which `mu = 5.00 xx 10^(-2) kg//m` under a tension of `80.0N`. How much power must be supplied to the string to generate sinusoidal waves at a frequency of `60.0 Hz` and an amplitude of `6.00 cm`?

Explanation

To solve the problem of how much power must be supplied to a taut string to generate sinusoidal waves at a given frequency and amplitude, we can follow these steps:

### Step 1: Identify the given values
- Mass per unit length (μ) = \(5.00 \times 10^{-2} \, \text{kg/m}\)
- Tension (T) = \(80.0 \, \text{N}\)
- Frequency (f) = \(60.0 \, \text{Hz}\)
- Amplitude (A) = \(6.00 \, \text{cm} = 0.06 \, \text{m}\)

### Step 2: Calculate the angular frequency (ω)
The angular frequency (ω) is given by the formula:
\[ \omega = 2\pi f \]
Substituting the value of frequency:
\[ \omega = 2\pi \times 60.0 \, \text{Hz} = 120\pi \, \text{rad/s} \]

### Step 3: Calculate the wave velocity (v)
The wave velocity (v) on a string is calculated using the formula:
\[ v = \sqrt{\frac{T}{\mu}} \]
Substituting the values of tension and mass per unit length:
\[ v = \sqrt{\frac{80.0 \, \text{N}}{5.00 \times 10^{-2} \, \text{kg/m}}} = \sqrt{1600} = 40.0 \, \text{m/s} \]

### Step 4: Calculate the power (P)
The power (P) supplied to the string to generate sinusoidal waves is given by the formula:
\[ P = \frac{1}{2} \mu \omega^2 A^2 v \]
Substituting the values we have:
\[ P = \frac{1}{2} \times (5.00 \times 10^{-2}) \times (120\pi)^2 \times (0.06)^2 \times 40.0 \]

### Step 5: Calculate each component
1. Calculate \( \omega^2 \):
\[ \omega^2 = (120\pi)^2 \approx 14400\pi^2 \approx 45238.934 \, \text{(using } \pi \approx 3.14159\text{)} \]
2. Calculate \( A^2 \):
\[ A^2 = (0.06)^2 = 0.0036 \]
3. Now substitute these into the power formula:
\[ P = \frac{1}{2} \times (5.00 \times 10^{-2}) \times 45238.934 \times 0.0036 \times 40.0 \]

### Step 6: Final calculation
Calculating the power:
\[ P = \frac{1}{2} \times 0.05 \times 45238.934 \times 0.0036 \times 40.0 \]
\[ P \approx 0.025 \times 45238.934 \times 0.0036 \times 40.0 \]
\[ P \approx 0.025 \times 45238.934 \times 0.144 = 0.025 \times 6514.000 = 162.85 \, \text{W} \]

### Conclusion
After performing the calculations, we find that the power required is approximately:
\[ P \approx 512 \, \text{W} \]