A piano wire has a length of 81 cm and a mass of 2.0 g. If its fundamental frequency is to be 394 Hz, it tension must be

Options

A. 0.32 N
B. 63 N
C. 130 N
D. 250 N

Answer

To find the tension in the piano wire given its length, mass, and fundamental frequency, we can follow these steps:

### Step 1: Understand the relationship between frequency, tension, and mass per unit length.
The fundamental frequency \( f \) of a string clamped at both ends is given by the formula:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
where:
- \( L \) is the length of the string,
- \( T \) is the tension in the string,
- \( \mu \) is the mass per unit length of the string.

### Step 2: Calculate the mass per unit length (\( \mu \)).
The mass per unit length \( \mu \) can be calculated as:
\[ \mu = \frac{m}{L} \]
where:
- \( m \) is the mass of the string,
- \( L \) is the length of the string.

Given:
- Length \( L = 81 \, \text{cm} = 0.81 \, \text{m} \)
- Mass \( m = 2.0 \, \text{g} = 2.0 \times 10^{-3} \, \text{kg} \)

Calculating \( \mu \):
\[ \mu = \frac{2.0 \times 10^{-3} \, \text{kg}}{0.81 \, \text{m}} \approx 2.469 \times 10^{-3} \, \text{kg/m} \]

### Step 3: Rearrange the formula to solve for tension \( T \).
Rearranging the formula for frequency gives:
\[ T = (2Lf)^2 \mu \]

### Step 4: Substitute the values into the tension formula.
Now substitute \( L \), \( f \), and \( \mu \) into the equation:
\[ T = (2 \times 0.81 \, \text{m} \times 394 \, \text{Hz})^2 \times (2.469 \times 10^{-3} \, \text{kg/m}) \]

### Step 5: Calculate the tension \( T \).
Calculating \( T \):
1. Calculate \( 2Lf \):
\[ 2Lf = 2 \times 0.81 \times 394 \approx 638.88 \, \text{m/s} \]
2. Square this result:
\[ (2Lf)^2 \approx (638.88)^2 \approx 408,000 \, \text{m}^2/\text{s}^2 \]
3. Now multiply by \( \mu \):
\[ T \approx 408,000 \times 2.469 \times 10^{-3} \approx 1005.9 \, \text{N} \]

### Final Answer:
The tension in the piano wire must be approximately \( 1005.9 \, \text{N} \).

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