Question

A string vibrates with a frequency of 200 Hz. When its length is doubled and tension is altered, it begins to vibrate with a frequency of 300 Hz. The ratio of the new tension to the original tension is

• A. 9:1
• B. 1:9
• C. 3:1
• D. 1:3

Hint:

The frequency of a vibrating string is proportional to the square root of the tension and inversely proportional to the square root of the length. A change in length or tension will significantly affect the frequency.

Answer 1

The Correct Option is A

The frequency \( f \) of a string vibrating under tension is given by the formula: \[ f \propto \sqrt{\frac{T}{L}} \] where: 
- \( f \) is the frequency, 
- \( T \) is the tension, 
- \( L \) is the length of the string. Let the initial frequency be \( f_1 = 200 \, \text{Hz} \), the initial length be \( L_1 \), and the initial tension be \( T_1 \). 
The frequency after the length is doubled and the tension is changed is \( f_2 = 300 \, \text{Hz} \), and the new length is \( L_2 = 2L_1 \). 
Using the formula for frequency, the ratio of the new frequency to the initial frequency is: \[ \frac{f_2}{f_1} = \sqrt{\frac{T_2}{T_1}} \cdot \sqrt{\frac{L_1}{L_2}} \] Since \( L_2 = 2L_1 \), the ratio becomes: \[ \frac{f_2}{f_1} = \sqrt{\frac{T_2}{T_1}} \cdot \frac{1}{\sqrt{2}} \] Now, squaring both sides: \[ \left( \frac{f_2}{f_1} \right)^2 = \frac{T_2}{T_1} \cdot \frac{1}{2} \] Substitute the known values: \[ \left( \frac{300}{200} \right)^2 = \frac{T_2}{T_1} \cdot \frac{1}{2} \] \[ \left( \frac{3}{2} \right)^2 = \frac{T_2}{T_1} \cdot \frac{1}{2} \] \[ \frac{9}{4} = \frac{T_2}{T_1} \cdot \frac{1}{2} \] \[ \frac{T_2}{T_1} = \frac{9}{2} \] 
Thus, the ratio of the new tension to the original tension is: \[ \text{(1) } 9:1 \]