Spring Oscillation: Kinetic Energy at a Given Displacement
Question:
A block oscillates on a very long horizontal spring. The amplitude of its oscillation is 10 cm.
How does the block’s kinetic energy K when it is 5 cm from equilibrium compare to its maximum kinetic energy Kmax?
- A. K < Kmax⁄2
- B. K = Kmax⁄2
- C. K = ¾ Kmax
- D. K = Kmax
Answer:
For a simple harmonic oscillator, the total mechanical energy E is conserved and given by:
E = ½ k A²
The kinetic energy K at any displacement x from equilibrium is:
K = ½ k (A² − x²)
And the maximum kinetic energy occurs when x = 0, so:
Kmax = ½ k A²
Step-by-step Calculation:
Given: A = 10 cm, x = 5 cm
K = ½ k (A² − x²) = ½ k (10² − 5²) = ½ k (100 − 25) = ½ k (75)
Kmax = ½ k × 100
⇒ K / Kmax = (½ k × 75) / (½ k × 100) = 75 / 100 = ¾
Conclusion:
When the block is 5 cm from equilibrium, its kinetic energy is ¾ of the maximum kinetic energy.
Hence, the correct answer is: C. K = ¾ Kmax