In order to increase the frequency of transverse oscillations of a stretched wire by 50%, its tension munst be increased by

Options

A. 0.5
B. 1
C. 1.25
D. 1.5

Answer

To solve the problem of how much the tension in a stretched wire must be increased to raise the frequency of transverse oscillations by 50%, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Frequency Relation**:
The frequency \( f \) of transverse oscillations in a stretched wire is given by the formula:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
where \( T \) is the tension in the wire, \( L \) is the length of the wire, and \( \mu \) is the mass per unit length of the wire.

2. **Determining the New Frequency**:
If we want to increase the frequency by 50%, the new frequency \( f_2 \) can be expressed as:
\[ f_2 = 1.5 f_1 \]
where \( f_1 \) is the initial frequency.

3. **Setting Up the Ratio of Frequencies**:
Using the frequency formula, we can set up the ratio of the new frequency to the old frequency:
\[ \frac{f_2}{f_1} = \frac{\frac{1}{2L} \sqrt{\frac{T_2}{\mu}}}{\frac{1}{2L} \sqrt{\frac{T_1}{\mu}}} = \sqrt{\frac{T_2}{T_1}} \]
Therefore, we have:
\[ 1.5 = \sqrt{\frac{T_2}{T_1}} \]

4. **Squaring Both Sides**:
Squaring both sides to eliminate the square root gives:
\[ (1.5)^2 = \frac{T_2}{T_1} \]
\[ 2.25 = \frac{T_2}{T_1} \]

5. **Finding the New Tension**:
Rearranging the equation gives:
\[ T_2 = 2.25 T_1 \]

6. **Calculating the Increase in Tension**:
The increase in tension \( \Delta T \) is given by:
\[ \Delta T = T_2 - T_1 = 2.25 T_1 - T_1 = 1.25 T_1 \]

7. **Finding the Factor of Increase**:
To find the factor by which the tension must be increased, we divide the increase by the original tension:
\[ \text{Factor of Increase} = \frac{\Delta T}{T_1} = \frac{1.25 T_1}{T_1} = 1.25 \]

### Conclusion:
Thus, the tension must be increased by a factor of **1.25** to raise the frequency of transverse oscillations by 50%.