The fundamental frequency of a wire of certain length is 400 Hz. When the length of the wire is decreased by 10 cm, without changing the tension in the wire, the frequency becomes 500 Hz. What was the original length of the wire ?

Answer

To solve the problem, we will use the relationship between the frequency of a vibrating wire and its length. The fundamental frequency \( f \) of a wire is given by the formula:

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

where:
- \( f \) is the frequency,
- \( L \) is the length of the wire,
- \( T \) is the tension in the wire,
- \( m \) is the mass per unit length of the wire.

### Step 1: Establish the relationship for the original frequency
Let the original length of the wire be \( L_1 \) and the original frequency be \( f_1 = 400 \, \text{Hz} \).

Using the formula, we have:

\[ f_1 = \frac{1}{2L_1} \sqrt{\frac{T}{m}} \]

### Step 2: Establish the relationship for the new frequency
When the length of the wire is decreased by 10 cm (which is \( 0.1 \, \text{m} \)), the new length \( L_2 \) becomes:

\[ L_2 = L_1 - 0.1 \]

The new frequency \( f_2 \) is given as \( 500 \, \text{Hz} \). Therefore, we can write:

\[ f_2 = \frac{1}{2L_2} \sqrt{\frac{T}{m}} \]

### Step 3: Set up the equations
Now we have two equations:

1. \( f_1 = \frac{1}{2L_1} \sqrt{\frac{T}{m}} \)
2. \( f_2 = \frac{1}{2(L_1 - 0.1)} \sqrt{\frac{T}{m}} \)

### Step 4: Divide the equations
Dividing the second equation by the first gives:

\[ \frac{f_2}{f_1} = \frac{L_1}{L_1 - 0.1} \]

Substituting the values of \( f_1 \) and \( f_2 \):

\[ \frac{500}{400} = \frac{L_1}{L_1 - 0.1} \]

### Step 5: Simplify the equation
This simplifies to:

\[ \frac{5}{4} = \frac{L_1}{L_1 - 0.1} \]

Cross-multiplying gives:

\[ 5(L_1 - 0.1) = 4L_1 \]

### Step 6: Solve for \( L_1 \)
Expanding the left side:

\[ 5L_1 - 0.5 = 4L_1 \]

Rearranging gives:

\[ 5L_1 - 4L_1 = 0.5 \]

Thus,

\[ L_1 = 0.5 \, \text{m} = 50 \, \text{cm} \]

### Final Answer
The original length of the wire is \( 50 \, \text{cm} \).

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