Oscillations

Physics · Class 11

Simulation · Physics · Class 11

Simple pendulum: length, g and small swings

"Simple pendulum: length, g and small swings" is a free, interactive Class 11 Physics simulation for NEET and JEE, built on the lesson "The simple pendulum" in Oscillations. Change the length and the angle it is pulled aside. Compare the small-angle period 2π√(L/g) with the true period of the swing. For small swings T = 2π√(L/g): four times the length gives twice the period, and the mass of the bob never appears. At large angles sin θ < θ, the restoring force is weaker than the formula assumes, and the swing takes longer. JEE Main asked 29 questions from this chapter across the 42 shifts Lumi has analysed (2017–2026).

From the lesson The simple pendulum in Oscillations. Change the values and watch what happens.

Simple pendulum: length, g and small swingsPhysics · Class 11

In JEE Main

42 shifts analysed

29 questions from Oscillations across the 42 JEE Main shifts Lumi has analysed (2017–2026).

See the past questions

The idea behind it

NCERT §13.8

  • Galileo is said to have timed a swinging church chandelier against his pulse and found its motion periodic. A stone on an inextensible thread about 100 cm long swings with a period of about two seconds.
  • The idealised simple pendulum is a small bob of mass m on a massless, inextensible string of length L, fixed at a rigid support and swinging in a vertical plane.
  • The bob feels just two forces: its weight mg and the string's tension T. Split the weight into mg cos θ, acting along the string, and mg sin θ, acting at right angles to it along the arc.
  • The bob moves on a circle of radius L, so it has a radial acceleration supplied by T − mg cos θ and a tangential acceleration supplied by mg sin θ. The radial force passes through the support and gives no torque about it.
  • The restoring torque about the support is τ = −L(mg sin θ), negative because it acts to reduce θ. With τ = Iα, this gives α = −(mgL/I) sin θ.
  • sin θ = θ − θ³/3! + θ⁵/5! − …, with θ in radians. For small θ, sin θ ≈ θ, and the two stay close even up to about 20°.
  • Then α = −(mgL/I)θ, which has the form of the SHM equation for angular displacement, with ω = √(mgL/I) and T = 2π√(I/mgL).
  • For a massless string I = mL², so T = 2π√(L/g). The period depends only on the length and on g; it does not depend on the mass of the bob, nor (for small swings) on the amplitude.
  • A seconds pendulum has T = 2 s (one second each way). From L = gT²/4π² with g = 9.8 m s⁻², L = 9.8 × 4/(4π²) ≈ 1 m.

Questions

3 answers

What does the "Simple pendulum: length, g and small swings" simulation show?
Change the length and the angle it is pulled aside. Compare the small-angle period 2π√(L/g) with the true period of the swing.
Does JEE Main ask questions from Oscillations?
Yes. JEE Main asked 29 questions from this chapter across the 42 shifts Lumi has analysed (2017–2026).
Is this simulation free?
Yes. It is free and runs in the browser on a phone or a laptop. It belongs to the lesson "The simple pendulum" in Oscillations (Class 11 Physics).

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