When the tension in a string is increased by 44%. the frequency increased by 10Hz the frequency of the string is

Options

A. 100 Hz
B. 200 Hz
C. 150 Hz
D. 50 Hz

Answer

To solve the problem, we need to determine the original frequency of the string when the tension is increased by 44%, resulting in a frequency increase of 10 Hz. Here’s a step-by-step solution:

### Step 1: Understand the relationship between tension and frequency
The frequency of a vibrating string is related to the tension in the string by the formula:
\[ f = n \frac{v}{2L} \]
where \( v = \sqrt{\frac{T}{\mu}} \). Thus, we can express frequency as:
\[ f = n \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
From this, we see that frequency \( f \) is directly proportional to the square root of the tension \( T \).

### Step 2: Set up the initial and final tensions
Let the initial tension be \( T_1 \) and the final tension after increasing it by 44% be \( T_2 \):
\[ T_1 = T \]
\[ T_2 = T_1 + 0.44 T_1 = 1.44 T_1 \]

### Step 3: Relate the frequencies
Let the initial frequency be \( f_1 \) and the final frequency be \( f_2 \):
\[ f_2 = f_1 + 10 \text{ Hz} \]
Since frequency is proportional to the square root of tension, we have:
\[ \frac{f_1}{f_2} = \sqrt{\frac{T_1}{T_2}} = \sqrt{\frac{T_1}{1.44 T_1}} = \sqrt{\frac{1}{1.44}} = \frac{1}{\sqrt{1.44}} = \frac{1}{1.2} \]

### Step 4: Set up the equation
Substituting \( f_2 = f_1 + 10 \) into the ratio gives:
\[ \frac{f_1}{f_1 + 10} = \frac{1}{1.2} \]

### Step 5: Cross-multiply and solve for \( f_1 \)
Cross-multiplying gives:
\[ 1.2 f_1 = f_1 + 10 \]
Rearranging this yields:
\[ 1.2 f_1 - f_1 = 10 \]
\[ 0.2 f_1 = 10 \]
\[ f_1 = \frac{10}{0.2} = 50 \text{ Hz} \]

### Conclusion
The original frequency of the string is \( 50 \text{ Hz} \).

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