A particle of mass 250 g executes a simple harmonic motion under a periodic force F=(-25 x)N. The particle attains a maximum speed of 4 m/s during its oscillation. The amplitude of the motion is ___________ cm.

Solution

Physics sir · 3 steps

Read the spring constant straight off the force, find ω, then use vₘₐₓ = Aω.

1. Spring constant. The force has the form F = -kx, so k = 25 N m⁻¹. The mass is m = 250 g = 0.25 kg.
2. Angular frequency. ω = √(k/m) = √(25/0.25) = √100 = 10 rad s⁻¹.

   vₘₐₓ = Aω ⇒ A = vₘₐₓ/ω = 4/10 = 0.4 m

3. So. The amplitude is 0.4 m, which is 40 cm.

The common slip: 4. Taking ω = k/m = 100 rad s⁻¹ without the square root gives an amplitude of 4 cm, not 40.

The idea to keep. In any SHM with F = -kx, the coefficient of x is k. Then ω = √(k/m), and maximum speed is always Aω, reached at the mean position.

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