Question

A sonometer wire gives frequency \( f_1 \) with tension \( T_1 \). If the tension is made 4 times greater, what is the new frequency? The frequency of a vibrating wire is related to the tension in the wire by the equation: \[ f \propto \sqrt{T} \] where:
\( f \) is the frequency,
\( T \) is the tension in the wire.

If the tension is increased by a factor of 4, how does the frequency change?

• A. \( f_2 = 2f_1 \)
• B. \( f_2 = 4f_1 \)
• C. \( f_2 = f_1 \)
• D. \( f_2 = \sqrt{4} \times f_1 \)

Hint:

The frequency of a sonometer wire is proportional to the square root of the tension. So, if the tension is increased by a factor, the frequency will increase by the square root of that factor.

Answer 1

The Correct Option is D

According to the formula for the frequency of a vibrating wire: \[ f \propto \sqrt{T} \] If the tension is increased by a factor of 4, the new frequency \( f_2 \) is related to the initial frequency \( f_1 \) by: \[ f_2 = f_1 \times \sqrt{\frac{T_2}{T_1}} = f_1 \times \sqrt{4} = 2f_1 \] Thus, the new frequency is twice the original frequency. The correct answer is Option (A): \( f_2 = 2f_1 \).