Question

When a mass of 200 gm hangs from the ceiling via spring in equilibrium, the extension in the spring is observed to be 2mm. Find angular frequency of its SHM and energy stored in equilibrium position respectively.

• A. \(\omega = 2 \, \text{rad/sec.}; U = 50\sqrt{2} \, \text{mJ}\)
• B. \(\omega = 50\sqrt{2} \, \text{rad/sec.}; U = 2 \, \text{mJ}\)
• C. \(\omega = 100\sqrt{2} \, \text{rad/sec.}; U = 4 \, \text{mJ}\)
• D. \(\omega = 25\sqrt{2} \, \text{rad/sec.}; U = 1 \, \text{mJ}\)

Hint:

To find the angular frequency in SHM, use the formula \(\omega = \sqrt{\frac{k}{m}}\), where \(k\) is the spring constant and \(m\) is the mass. The energy stored in a spring is given by \(U = \frac{1}{2}kx^2\).

Answer 1

The Correct Option is B

At equilibrium: \[ k x_0 = mg \] Given that \( k = 1000 \, \text{N/m} \) and \( x_0 = 2 \times 10^{-3} \, \text{m} \): \[ k \times 2 \times 10^{-3} = 200 \times 10^{-3} \times 10 \] \[ k = 1000 \, \text{N/m} \] Now, for the angular frequency \( \omega \): \[ \omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{1000}{200 \times 10^{-3}}} = \sqrt{\frac{1000000}{200}} = \sqrt{5000} \] \[ \omega = 50 \sqrt{2} \] The potential energy \( U \) is given by: \[ U = \frac{1}{2} k x_0^2 \] \[ U = \frac{1}{2} \times 1000 \times (2 \times 10^{-3})^2 = 2 \times 10^{-3} \, \text{J} \] Thus, the potential energy \( U \) is: \[ U = 2 \, \text{mJ} \]