Simple Pendulum Simulator — find g सरल लोलक
JEE Main / NEET Physics Unit 10: Oscillations and Waves · Unit 20: Experimental Skills · NCERT Class 11 Physics, Oscillations · SLH original
⚠️ Simulated values from the model below — not real lab measurements.
Your readings (trials)
| # | L (m) | θ₀ (°) | T (s) | T² (s²) |
|---|
Guided experiment — step by step
- Set θ₀ = 10°. Choose L = 0.4 m, wait for 1–2 swings, press "Record reading". Repeat for L = 0.8, 1.2, 1.6 and 2.0 m.
- Look at the table: does T² grow in proportion to L? The slope and your value of g appear below the table.
- Now keep L = 1 m and change θ₀ from 5° to 60°. Compare the measured period with 2π√(L/g) — the small-angle formula breaks down at large angles.
- Change g to Moon and repeat one reading. Is the period longer or shorter?
Formula, units & assumptions
For small angles: T = 2π √(L / g) → T² = (4π² / g) · L, a straight line through the origin with slope 4π²/g, so g = 4π² / slope.
T does not depend on the mass of the bob. For larger angles the true period is a little longer: T ≈ T₀(1 + θ₀²/16), θ₀ in radians.
Units: length in metres (m), time in seconds (s), speed in m/s, g in m/s², angles in degrees on screen (radians inside formulas).
Assumptions & limits of the model: Point mass bob · massless, non-stretching string · no air drag or friction (the swing never dies down) · rigid support. The "measured" period is timed from the simulation, so it is a simulated value, not a real lab measurement.
Exam connection — questions
Concept check (MCQ)
1. The length of a simple pendulum is made 4 times. Its time period becomes:
2. For small oscillations, the time period of a simple pendulum does NOT depend on:
3. The graph of T² against L for a simple pendulum is:
Numerical answer (type a number)
A simple pendulum has length 1 m at a place where g = π² m/s². Find its time period in seconds.
A pendulum has a period of 2 s on Earth. Find its period (in s, one decimal place) on the Moon, where g is one-sixth of Earth's.