Question

In an experiment with sonometer when a mass of 180 g is attached to the string, it vibrates with fundamental frequency of 30 Hz. When a mass m is attached, the string vibrates with fundamental frequency of 50 Hz. The value of m is ______g.

Answer 1

Correct Answer: 500

The fundamental frequency of a vibrating string is given by: \[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}, \] where \( f \) is the frequency, \( L \) is the length of the string, \( T \) is the tension in the string, and \( \mu \) is the linear mass density. The ratio of frequencies for the two cases is: \[ \frac{f_2}{f_1} = \sqrt{\frac{T_2}{T_1}}, \] where \( T_1 = 180g \, \text{and} \, T_2 = mg \). Substitute \( f_2 = 50 \, \text{Hz}, f_1 = 30 \, \text{Hz} \): \[ \frac{50}{30} = \sqrt{\frac{mg}{180g}}. \] Simplify: \[ \left(\frac{50}{30}\right)^2 = \frac{m}{180}. \] \[ \frac{25}{9} = \frac{m}{180}. \] Solve for \( m \): \[ m = \frac{25}{9} \cdot 180 = 500 \, \text{g}. \] Thus, the value of \( m \) is \( \boxed{500 \, \text{g}} \).