The amplitude of a particle executing SHM is 3 cm. The displacement at which its kinetic energy will be 25 % more than the potential energy is: __________ cm

Solution

Physics sir · 4 steps

Write KE and PE in terms of x, set KE equal to 1.25 times PE, and the spring constant cancels.

1. Energies at x. At displacement x, the potential energy is 1/2kx² and the kinetic energy is 1/2k(A² - x²). Together they always add up to 1/2kA².
2. Set the condition. KE being 25% more than PE means KE = 1.25 PE. The factor 1/2k cancels from both sides.

   A² - x² = 1.25 x² ⇒ A² = 2.25 x²

3. Put in numbers. With A = 3 cm, x² = 9/2.25 = 4 cm².
4. So. x = 1.5 times smaller than A, that is x = 3/1.5 = 2 cm.

The common slip: 2.24. Putting the 1.25 on the wrong side, PE = 1.25 KE, gives x = √5 ≈ 2.24 cm, not 2.

The idea to keep. In SHM, PE goes as x² and KE as A² - x². Any energy ratio between them turns into a simple equation in x² with k cancelling out.

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