A wire 0.5 m long vibrates 100 times a second. If the length of the wire is shortened to 0.4 in and the stretching force is increased to 4 times its original value what will be the new frequency?

Answer

To find the new frequency of the wire after it has been shortened and the stretching force increased, we can follow these steps:

### Step 1: Identify the given values
- Original length of the wire, \( L_1 = 0.5 \, \text{m} \)
- Original frequency, \( f_1 = 100 \, \text{Hz} \)
- New length of the wire, \( L_2 = 0.4 \, \text{m} \)
- The stretching force is increased to 4 times its original value, so \( T_2 = 4T_1 \)

### Step 2: Write the formula for frequency
The frequency of a vibrating wire is given by the formula:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{m}} \]
where \( L \) is the length of the wire, \( T \) is the tension (stretching force), and \( m \) is the mass per unit length of the wire.

### Step 3: Write the equations for the original and new frequency
For the original frequency \( f_1 \):
\[ f_1 = \frac{1}{2L_1} \sqrt{\frac{T_1}{m}} \]

For the new frequency \( f_2 \):
\[ f_2 = \frac{1}{2L_2} \sqrt{\frac{T_2}{m}} \]

### Step 4: Substitute the values into the equations
Substituting \( L_1 \), \( L_2 \), \( T_2 = 4T_1 \) into the equations:
\[ f_1 = \frac{1}{2 \times 0.5} \sqrt{\frac{T_1}{m}} = \frac{1}{1} \sqrt{\frac{T_1}{m}} = \sqrt{\frac{T_1}{m}} \]

\[ f_2 = \frac{1}{2 \times 0.4} \sqrt{\frac{4T_1}{m}} = \frac{1}{0.8} \sqrt{\frac{4T_1}{m}} = \frac{1}{0.8} \cdot 2 \sqrt{\frac{T_1}{m}} = \frac{2}{0.8} \sqrt{\frac{T_1}{m}} = 2.5 \sqrt{\frac{T_1}{m}} \]

### Step 5: Relate \( f_2 \) to \( f_1 \)
From the original frequency \( f_1 \):
\[ f_1 = \sqrt{\frac{T_1}{m}} \]
Thus, we can express \( f_2 \) in terms of \( f_1 \):
\[ f_2 = 2.5 f_1 \]

### Step 6: Calculate the new frequency
Now substituting \( f_1 = 100 \, \text{Hz} \):
\[ f_2 = 2.5 \times 100 = 250 \, \text{Hz} \]

### Final Answer
The new frequency \( f_2 \) is \( 250 \, \text{Hz} \).

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