Question

A particle is executing simple harmonic motion with an amplitude of 10 cm. If the kinetic energy of the particle at a distance of 6 cm from the mean position is 100 J, then the kinetic energy of the particle at a distance of 2 cm from the mean position is

• A. 225 J
• B. 300 J
• C. 150 J
• D. 75 J

Hint:

In SHM, always use $KE = E \left(1 - \frac{x^2}{A^2}\right)$ and $PE = E \frac{x^2}{A^2}$ to find energies at any displacement.

Answer 1

The Correct Option is A

1. Total energy in SHM: $E = \frac{1}{2} k A^2 = KE + PE$.
2. At $x = 6$ cm, $KE = 100$ J, $PE = \frac{1}{2} k (A^2 - x^2) = E - KE$
3. Compute $E$: $E = KE + PE = KE + (E - KE)$ etc.
4. At $x = 2$ cm, $KE = E - \frac{1}{2} k x^2 = 225$ J.