Question

A block kept on a frictionless horizontal surface is connected to one end of a horizontal spring of constant \( 100 \, \text{Nm}^{-1} \), whose other end is fixed to a rigid vertical wall. Initially, the block is at its equilibrium position. The block is pulled to a distance of \( 8 \) cm and released. The kinetic energy of the block when it is at a distance of \( 3 \) cm from the mean position is?

• A. \( 0.65 \, J \)
• B. \( 0.325 \, J \)
• C. \( 0.275 \, J \)
• D. \( 0.5 \, J \)

Hint:

In simple harmonic motion, the total mechanical energy remains constant, and kinetic energy at any position can be found using \( KE = E - PE \), where \( PE = \frac{1}{2} k x^2 \).

Answer 1

The Correct Option is C

Step 1: Understanding energy conservation in simple harmonic motion 
In simple harmonic motion (SHM), the total mechanical energy is conserved: \[ E = \frac{1}{2} k A^2 \] where: - \( k = 100 \, \text{Nm}^{-1} \) (spring constant), - \( A = 8 \) cm = \( 0.08 \) m (amplitude of oscillation). Thus, the total energy of the system is: \[ E = \frac{1}{2} \times 100 \times (0.08)^2 \] \[ = 50 \times 0.0064 = 0.32 \, J \] Step 2: Find the kinetic energy at \( x = 3 \) cm 
The total mechanical energy in SHM is the sum of kinetic energy \( KE \) and potential energy \( PE \): \[ E = KE + PE \] Potential energy at displacement \( x = 3 \) cm = \( 0.03 \) m is: \[ PE = \frac{1}{2} k x^2 = \frac{1}{2} \times 100 \times (0.03)^2 \] \[ = 50 \times 0.0009 = 0.045 \, J \] Step 3: Compute kinetic energy 
\[ KE = E - PE \] \[ KE = 0.32 - 0.045 \] \[ KE = 0.275 \, J \] Thus, the kinetic energy of the block when it is at a distance of \( 3 \) cm from the mean position is \( 0.275 \) J.