An object of mass 0.2 kg executes simple harmonic motion along x axis with frequency of (25/π) Hz. At the position x=0.04 m the object has kinetic energy 0.5 J and potential energy 0.4 J. The amplitude of oscillation is ________ cm.

Solution

Physics sir · 3 steps

Add KE and PE to get the total energy, then set it equal to 1/2kA² and solve for A.

1. Find the spring constant. ω = 2π f = 2π × 25/π = 50 rad s⁻¹. So k = mω² = 0.2 × 2500 = 500 N m⁻¹. Check: 1/2kx² = 1/2 × 500 × (0.04)² = 0.4 J, which matches the given potential energy.
2. Total energy. In SHM the total energy stays the same at every position. Here it is E = 0.5 + 0.4 = 0.9 J. At the extreme position all of it is potential energy.

   1/2kA² = E ⇒ A² = (2 × 0.9)/500 = 0.0036 m²

3. So. A = √0.0036 = 0.06 m, which is 6 cm.

The common slip: 0.06. Stopping at A = 0.06 in metres gives 0.06, but the blank asks for centimetres, so multiply by 100.

The idea to keep. In SHM, KE + PE at any point equals 1/2kA². Since PE/E = x²/A², you can even skip k: 0.4/0.9 = (0.04)²/A² gives A in one line.

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