Tip: drag the bob with your mouse or finger to set the angle, then let go to start swinging. Press the Space key to pause or resume. Origin O is the lowest point of the swing.
Angle vs time graph
A smooth wave shows simple harmonic motion. The gap between two peaks is the time period.
Results
Live values
KE + PE stays constant, because there is no air friction.
Theory
A simple pendulum is a small heavy bob hanging from a light string of length L. When it is pulled aside and released, gravity pulls it back towards the lowest point and it swings to and fro. This motion is called oscillation.
For small angles (up to about 15°) the motion is simple harmonic and the time period is:
The bigger the starting angle, the longer the real time period gets. The "Exact time period" in the results shows this.
Forces on the bob (turn on "Show forces" to see them):
Tension in the string: T = m v² / L + m g cosθ
Restoring force (pulls the bob back): F = m g sinθ
Energy and speed (h is the height above the lowest point):
KE = ½ m v²
Speed at angle θ: v = √(2 g L (cosθ − cosθ₀))
Maximum speed (at the lowest point): v = √(2 g L (1 − cosθ₀))
Key points:
- The time period does not depend on the mass of the bob.
- The time period is proportional to √L. If L becomes 4 times, T becomes 2 times.
- The time period is inversely proportional to √g. On the Moon the pendulum swings more slowly.
- A pendulum with T = 2 s is called a seconds pendulum.
- The speed is maximum at the lowest point and zero at the two extreme points.
- The tension is largest at the lowest point and smallest at the extreme points.
- With air friction the swings slowly become smaller and the energy is lost as heat.
Practice questions
Q1. Find the length of a seconds pendulum (T = 2 s) on Earth. Take g = 9.8 m/s².
L = g T² / (4π²) = 9.8 × 4 / 39.48 = 0.99 m (about 1 m)
Q2. The length of a pendulum is made 4 times. What happens to its time period?
T is proportional to √L, so T becomes √4 = 2 times. The time period doubles.
Q3. A pendulum of length 0.5 m is taken to the Moon (g = 1.62 m/s²). Find its time period.
T = 2π √(0.5 / 1.62) = 2π × 0.556 = 3.49 s
Q4. A pendulum of length 1 m is released from 60°. Find its maximum speed. Take g = 9.8 m/s².
v = √(2 g L (1 − cos60°)) = √(2 × 9.8 × 1 × 0.5) = √9.8 = 3.13 m/s
Q5. For the pendulum in Q4 (mass 1 kg), find the tension in the string at the lowest point.
T = m g + m v² / L = 9.8 + 1 × 9.8 / 1 = 19.6 N
Tip: Put these values in the controls above and check your answers in the simulation.