Question

When a body of mass 8 kg is attached to a spring balance, the reading of the balance is 20 cm. Instead of 8 kg, if another body of mass M is suspended from the spring balance and is made to oscillate vertically, the time period of oscillation is \( \frac{\pi}{5} \) s, then the value of M is (Acceleration due to gravity = 10 ms\(^{-2} \))

• A. \( 4 \) kg
• B. \( 6 \) kg
• C. \( 8 \) kg
• D. \( 9 \) kg

Hint:

First, use the static equilibrium condition of the spring balance to find the spring constant \( k \). Then, use the formula for the time period of vertical oscillations of a mass-spring system to relate the mass \( M \), the spring constant \( k \), and the given time period \( T \). Solve for \( M \). Remember to convert units to be consistent (e.g., cm to m).

Answer 1

The Correct Option is A

When a mass \( m \) is attached to a spring balance, the spring extends until the spring force balances the gravitational force: \( kx = mg \), where \( k \) is the spring constant and \( x \) is the extension.
In the first case, mass \( m_1 = 8 \) kg and extension \( x_1 = 20 \) cm \( = 0.
2 \) m.
\( k(0.
2) = 8(10) \) \( 0.
2k = 80 \) \( k = \frac{80}{0.
2} = 400 \) N/m When a mass \( M \) is suspended and oscillates vertically, the time period of oscillation for a mass-spring system is given by \( T = 2\pi \sqrt{\frac{M}{k}} \).
Given time period \( T = \frac{\pi}{5} \) s.
\( \frac{\pi}{5} = 2\pi \sqrt{\frac{M}{400}} \) Divide both sides by \( \pi \): \( \frac{1}{5} = 2 \sqrt{\frac{M}{400}} \) Divide both sides by 2: \( \frac{1}{10} = \sqrt{\frac{M}{400}} \) Square both sides: \( \left(\frac{1}{10}\right)^2 = \frac{M}{400} \) \( \frac{1}{100} = \frac{M}{400} \) Solve for \( M \): \( M = \frac{400}{100} = 4 \) kg The value of M is 4 kg.