The tension in a piano wire is `10 N`. The tension ina piano wire to produce a node of double frequency is

Options

A. `20 N`
B. `40 N`
C. `10 N`
D. ` 120 N`

Explanation

To find the tension in a piano wire that produces a note of double frequency, we can use the relationship between frequency, tension, and mass per unit length. The formula for frequency \( f \) of a vibrating string (or wire) is given by:

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

where:
- \( f \) = frequency of the wire
- \( L \) = length of the wire
- \( T \) = tension in the wire
- \( \mu \) = mass per unit length of the wire

### Step 1: Establish the initial frequency
Let the initial tension be \( T_1 = 10 \, \text{N} \) and the initial frequency be \( f_1 \).

Using the formula for frequency, we have:

\[ f_1 = \frac{1}{2L} \sqrt{\frac{T_1}{\mu}} = \frac{1}{2L} \sqrt{\frac{10}{\mu}} \]

### Step 2: Determine the frequency for double the initial frequency
We want to find the tension \( T_2 \) that produces a frequency \( f_2 = 2f_1 \).

Using the frequency formula again for the new tension:

\[ f_2 = \frac{1}{2L} \sqrt{\frac{T_2}{\mu}} \]

### Step 3: Set up the equation for double frequency
Since \( f_2 = 2f_1 \), we can write:

\[ 2f_1 = \frac{1}{2L} \sqrt{\frac{T_2}{\mu}} \]

### Step 4: Substitute \( f_1 \) into the equation
Substituting \( f_1 \) from Step 1 into the equation gives:

\[ 2 \left( \frac{1}{2L} \sqrt{\frac{10}{\mu}} \right) = \frac{1}{2L} \sqrt{\frac{T_2}{\mu}} \]

### Step 5: Simplify the equation
We can cancel \( \frac{1}{2L} \) from both sides:

\[ 2 \sqrt{\frac{10}{\mu}} = \sqrt{\frac{T_2}{\mu}} \]

### Step 6: Square both sides
Squaring both sides gives:

\[ 4 \cdot \frac{10}{\mu} = \frac{T_2}{\mu} \]

### Step 7: Solve for \( T_2 \)
Multiplying both sides by \( \mu \):

\[ T_2 = 40 \, \text{N} \]

### Conclusion
The tension in the piano wire to produce a note of double frequency is \( T_2 = 40 \, \text{N} \).

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