Question

Two small particles of mass 20 gm and 30 gm are connected with a rigid massless rod of length of 10 cm. The system's center of mass is suspended by a steel wire of torsional spring constant \(c=1.2\times 10^{-8}\) N.m/rad. If the system is slightly rotated in its plane and the angular frequency of its oscillations in rad/sec is \(n\times 10^{-3}\), then write the value of n.

Answer 1

Correct Answer: 10

Moment of Inertia and Angular Frequency Calculation 

The reduced mass \( m_{eq} \) is given by: \[ \frac{m_1 m_2}{m_1 + m_2} = \frac{(20)(30)}{20 + 30} = 12 \, \text{gm} = 12 \times 10^{-3} \, \text{kg} \]

The moment of inertia \( I_{cm} \) is: \[ I_{cm} = m_{eq} r^2 = (12 \times 10^{-3}) (0.1)^2 = 12 \times 10^{-5} \, \text{kg} \cdot \text{m}^2 \]

The time period \( T \) is given by: \[ T = 2\pi \sqrt{\frac{I_{cm}}{C}} = \omega_n = \sqrt{\frac{C}{I_{cm}}} = \sqrt{\frac{1.2 \times 10^{-8}}{12 \times 10^{-5}}} \]

Solving for \( \omega_n \): \[ \omega_n = 10 \times 10^{-3} \, \text{rad/sec} \]

The angular frequency of its oscillations is \( n \times 10^{-3} \, \text{rad/sec} \), where: \[ n = 10 \]

Answer: The value of \( n \) is \( \boxed{10} \).