The position, velocity and acceleration of a particle executing simple harmonic motion are found to have magnitudes of 4 m, 2 ms⁻¹ and 16 ms⁻² at a certain instant. The amplitude of the motion is √x, m where x is _________.

Solution

Physics sir · 3 steps

Use the acceleration to get ω first, then the velocity relation hands you the amplitude.

1. Find omega. In SHM the acceleration magnitude is a = ω² x. So 16 = ω² × 4, which gives ω² = 4 and ω = 2 rad s⁻¹.

   v = ω√(A² - x²)

2. Put in numbers. 2 = 2√(A² - 16), so √(A² - 16) = 1. Squaring gives A² - 16 = 1.
3. So. A² = 17, so A = √17 m and x is 17.

The common slip: 16.25. Taking ω = a/x = 4 instead of ω² = a/x gives A² = 16 + 4/16 = 16.25, not 17.

The idea to keep. Position and acceleration together give ω through a = ω² x. Then v = ω√(A² - x²) links speed, position and amplitude.

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