If the tension in the string is increased by 5 kg wt, the frequency of the fundamental tone increases in the ratio 2:3.What was the initial tension in the string ?

Answer

To solve the problem step by step, we can follow these calculations:

### Step 1: Understand the relationship between frequency and tension
The frequency of the fundamental tone of a string is given by the formula:
\[ F = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
where \( F \) is the frequency, \( T \) is the tension in the string, \( L \) is the length of the string, and \( \mu \) is the linear mass density of the string.

### Step 2: Establish the initial and final tensions
Let the initial tension in the string be \( T \). When the tension is increased by 5 kg weight (which is equivalent to \( 5 \times 9.81 \, \text{N} = 49.05 \, \text{N} \)), the new tension becomes:
\[ T' = T + 50 \, \text{N} \]

### Step 3: Set up the frequency ratio
According to the problem, the frequency increases in the ratio \( \frac{F'}{F} = \frac{2}{3} \). Therefore, we can express the frequencies in terms of the tensions:
\[ \frac{F'}{F} = \frac{\sqrt{T'}}{\sqrt{T}} = \frac{2}{3} \]

### Step 4: Square both sides to eliminate the square roots
Squaring both sides gives:
\[ \frac{T'}{T} = \left(\frac{2}{3}\right)^2 = \frac{4}{9} \]

### Step 5: Substitute the expression for \( T' \)
Substituting \( T' = T + 50 \) into the equation:
\[ \frac{T + 50}{T} = \frac{4}{9} \]

### Step 6: Cross-multiply to solve for \( T \)
Cross-multiplying gives:
\[ 9(T + 50) = 4T \]
Expanding this gives:
\[ 9T + 450 = 4T \]

### Step 7: Rearranging the equation
Rearranging the equation to isolate \( T \):
\[ 9T - 4T = -450 \]
\[ 5T = -450 \]

### Step 8: Solve for \( T \)
Dividing both sides by 5:
\[ T = \frac{450}{5} = 90 \, \text{N} \]

### Conclusion
The initial tension in the string is \( 90 \, \text{N} \).

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