Question

Two identical strings of length \( l \) and \( 2l \) vibrate with fundamental frequencies \( N \) hertz and \( 1.5N \) hertz, respectively. The ratio of tensions for smaller length to larger length is

• A. \( 9 : 1 \)
• B. \( 3 : 1 \)
• C. \( 1 : 9 \)
• D. \( 1 : 3 \)

Hint:

The tension in a vibrating string is proportional to the square of the frequency and inversely proportional to the length.

Answer 1

The Correct Option is C

Step 1: Formula for frequency of a vibrating string.
The fundamental frequency \( f \) of a vibrating string is given by: \[ f = \frac{1}{2l} \sqrt{\frac{T}{\mu}} \] where \( l \) is the length of the string, \( T \) is the tension, and \( \mu \) is the mass per unit length.
Step 2: Frequency ratio.
Since the strings are identical, we can relate the frequencies and tensions using the formula. The frequency is inversely proportional to the square root of the length and directly proportional to the square root of the tension. Therefore, the ratio of tensions is: \[ \frac{T_1}{T_2} = \left(\frac{f_2}{f_1}\right)^2 = \left(\frac{1.5}{1}\right)^2 = 9 \]
Step 3: Conclusion.
Thus, the ratio of tensions is \( 1 : 9 \), and the correct answer is (C).