A simple pendulum with length 100 cm and bob of mass 250 g is executing S.H.M. of amplitude 10 cm. The maximum tension in the string is found to be x/40 N. The value of x is ___________.

Solution

Physics sir · 4 steps

Tension peaks at the lowest point, where the bob is fastest: add mg and the centripetal force there.

1. Where tension peaks. At the mean position the bob moves fastest, and gravity points straight down along the string. So the string must hold up the weight and also supply the centripetal force.

   Tₘₐₓ = mg + (m vₘₐₓ²)/L

2. Maximum speed. In SHM, vₘₐₓ = Aω with ω = √(g/L). Taking g = 10 m s⁻², vₘₐₓ² = gA²/L = (10 × (0.1)²)/1 = 0.1 m² s⁻².
3. Put in numbers. With m = 0.25 kg and L = 1 m, mg = 2.5 N and (m vₘₐₓ²)/L = 0.25 × 0.1 = 0.025 N. So Tₘₐₓ = 2.525 N.
4. So. x/40 = 2.525 gives x = 40 × 2.525 = 101.

The idea to keep. For a pendulum, the maximum tension is at the lowest point. For small oscillations it is T = mg(1 + A²/L²), because vₘₐₓ = A√(g/L).

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