When the stretching forec of a wire increased by 2.5 kg, the frequency of the note emitted is changed in ration 2/3.Calculate the original force

Answer

To solve the problem step by step, we will use the relationship between the frequency of a vibrating wire and the tension in the wire.

### Step 1: Understand the relationship between frequency and tension
The frequency of a wire under tension is given by the formula:
\[ f \propto \sqrt{T} \]
where \( f \) is the frequency and \( T \) is the tension in the wire. This means that the frequency is proportional to the square root of the tension.

### Step 2: Set up the initial and final conditions
Let the original tension (force) in the wire be \( T \) (in kg). When the tension is increased by 2.5 kg, the new tension becomes:
\[ T' = T + 2.5 \]

### Step 3: Write the ratio of frequencies
According to the problem, the ratio of the frequencies when the tension is changed is given as:
\[ \frac{f'}{f} = \frac{3}{2} \]
This implies:
\[ f' = \frac{3}{2} f \]

### Step 4: Relate the frequencies to tensions
Using the relationship between frequency and tension, we can express this as:
\[ \frac{f'}{f} = \frac{\sqrt{T + 2.5}}{\sqrt{T}} \]
Setting this equal to the given ratio:
\[ \frac{\sqrt{T + 2.5}}{\sqrt{T}} = \frac{3}{2} \]

### Step 5: Square both sides to eliminate the square roots
Squaring both sides gives:
\[ \frac{T + 2.5}{T} = \left(\frac{3}{2}\right)^2 \]
\[ \frac{T + 2.5}{T} = \frac{9}{4} \]

### Step 6: Cross-multiply to solve for T
Cross-multiplying gives:
\[ 4(T + 2.5) = 9T \]
Expanding this:
\[ 4T + 10 = 9T \]

### Step 7: Rearrange to isolate T
Rearranging the equation gives:
\[ 10 = 9T - 4T \]
\[ 10 = 5T \]
Dividing both sides by 5:
\[ T = 2 \text{ kg} \]

### Final Answer
The original force (tension) in the wire is:
\[ \boxed{2 \text{ kg}} \]

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