A particle of mass 0.50 kg executes simple harmonic motion under force F=-50(Nm⁻¹) x. The time period of oscillation is x/35 s. The value of x is _________.
(Given π=22/7)

Solution

Physics sir · 3 steps

Read the spring constant straight off the force law, then T = 2π√(m/k) does the rest.

1. Find k. SHM always has the form F = -kx. Matching this with F = -50x gives k = 50 N m⁻¹. Here x in the force law is the displacement, not the blank you have to fill.

   T = 2π√(m/k)

2. Put in numbers. √(0.50/50) = √0.01 = 0.1, so T = 2 × 22/7 × 0.1 = 4.4/7 = 22/35 s.
3. So. Comparing 22/35 with x/35 gives x = 22.

The common slip: 2200. Writing T = 2π√(k/m) with the ratio upside down gives T = 440/7 s, so x comes out as 2200 instead of 22.

The idea to keep. Whenever a force is written as F = -(constant) x, that constant is your k. Then T = 2π√(m/k): mass on top, stiffness at the bottom.

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