A particle performs simple harmonic motion with amplitude A. Its speed is increased to three times at an instant when its displacement is (2 A)/3. The new amplitude of motion is (n A)/3. The value of n is ___________.

Solution

Physics sir · 5 steps

The kick changes the speed, not the position or ω, so use v² = ω²(A² - x²) before and after.

1. What stays fixed. The sudden push changes only the speed. The displacement at that instant is still 2A/3, and ω is set by the system, so it does not change either.

   v² = ω²(A² - x²)

2. Speed before. At x = 2A/3: v² = ω²(A² - 4A²/9) = (5ω² A²)/9.
3. Speed after. The speed becomes 3v, so v'² = 9v² = 5ω² A². The new amplitude A' must satisfy v'² = ω²(A'² - 4A²/9).
4. Solve for A'. A'² = 4A²/9 + 5A² = 49A²/9, so A' = 7A/3.
5. So. Comparing with nA/3 gives n = 7.

The common slip: 9. Assuming the amplitude triples along with the speed gives 3A, that is n = 9. The amplitude depends on both x and v, so it does not scale with the speed alone.

The idea to keep. When a particle in SHM gets a sudden kick, x and ω stay the same at that instant. Find the new speed, then get the new amplitude from A'² = x² + v'²/ω².

Answer from the NTA final answer key. Solution by PrepWiser.