A piano wire weighing 6.00 g and having a length of 90.0 cm emits a fundamental frequency corresponding to the "Middle C" (v = 261.63 Hz). Find the tension in the wire.

Explanation

To find the tension in the piano wire, we can follow these steps:

### Step 1: Convert the mass of the wire to kilograms
The mass of the wire is given as 6.00 g. We need to convert this to kilograms since the standard unit of mass in physics is kilograms.

\[ \text{Mass (m)} = 6.00 \, \text{g} = \frac{6.00}{1000} \, \text{kg} = 0.006 \, \text{kg} \]

### Step 2: Convert the length of the wire to meters
The length of the wire is given as 90.0 cm. We need to convert this to meters.

\[ \text{Length (L)} = 90.0 \, \text{cm} = \frac{90.0}{100} \, \text{m} = 0.9 \, \text{m} \]

### Step 3: Calculate the mass per unit length (μ)
Mass per unit length (μ) is calculated using the formula:

\[ \mu = \frac{m}{L} \]

Substituting the values we have:

\[ \mu = \frac{0.006 \, \text{kg}}{0.9 \, \text{m}} = \frac{0.006}{0.9} \, \text{kg/m} = 0.00667 \, \text{kg/m} \]

### Step 4: Use the formula for the fundamental frequency
The fundamental frequency (f) of a wire is given by the formula:

\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]

Where:
- \( f \) is the frequency (261.63 Hz)
- \( L \) is the length of the wire (0.9 m)
- \( T \) is the tension in the wire (which we need to find)
- \( \mu \) is the mass per unit length (0.00667 kg/m)

### Step 5: Rearrange the formula to solve for tension (T)
We can rearrange the formula to solve for T:

\[ T = (2Lf)^2 \mu \]

### Step 6: Substitute the values into the equation
Now we can substitute the known values into the equation:

\[ T = (2 \times 0.9 \, \text{m} \times 261.63 \, \text{Hz})^2 \times 0.00667 \, \text{kg/m} \]

Calculating the term inside the parentheses first:

\[ 2 \times 0.9 \times 261.63 = 471.03 \]

Now squaring this value:

\[ (471.03)^2 = 221,867.76 \]

Now, substituting back to find T:

\[ T = 221,867.76 \times 0.00667 \approx 1470.00 \, \text{N} \]

### Final Answer
The tension in the wire is approximately:

\[ T \approx 1470.00 \, \text{N} \]