Simple Pendulum Physics Simulation
Explore the relationship between pendulum length, period, and gravitational acceleration
Simulation
Theory
Quiz
Interactive Pendulum
20.0 cm
10 cm
30 cm
0.1 cm
0.05 cm
0.5 cm
90 s
80 s
100 s
1 s
0.1 s
2 s
Results
Period (T) = t/n = 90/100 = 0.9 s
g = 4π²L/T² = 4π²×0.20/(0.9)² = 9.74 m/s²
Percentage error in g: 3.2%
Error contributions:
- From length measurement (ΔL/L): 0.5%
- From time measurement (2×Δt/t): 2.2%
Δg/g = ΔL/L + 2(Δt/t) = 0.1/20.0 + 2(1/90) = 0.032
Pendulum Theory
Dimensional Analysis
The period T of a simple pendulum depends on:
- Length of string (L)
- Acceleration due to gravity (g)
- Mass of bob (m)
We assume the relationship:
T = k Lˣ gʸ mᶻ
Where k is a dimensionless constant. Writing the dimensional equation:
[M⁰L⁰T¹] = [L]ˣ [LT⁻²]ʸ [M]ᶻ
Equating dimensions:
For M: 0 = z
For L: 0 = x + y
For T: 1 = -2y
For L: 0 = x + y
For T: 1 = -2y
Solving gives:
y = -½, x = ½, z = 0
Thus the period is:
T = k √(L/g)
Experiment shows k = 2π for small angles, giving:
T = 2π √(L/g)
Error Analysis
The calculated value of g depends on measurements of L and T:
g = 4π² L/T²
The relative error in g is:
Δg/g = ΔL/L + 2(ΔT/T)
Since T = t/n (where n is number of oscillations timed):
ΔT/T = Δt/t
Test Your Understanding
1. The period of a simple pendulum depends on:
Correct! Dimensional analysis shows the period depends only on length (L) and gravitational acceleration (g). The mass (m) has exponent 0 in the dimensional analysis.
2. If you double the length of a pendulum, the period becomes:
Correct! Since T ∝ √L, doubling the length increases the period by √2 ≈ 1.414 times. Try adjusting the length in the simulation to verify this relationship.
3. Which measurement typically contributes more to error in g?
Correct! The time measurement error is multiplied by 2 in the error formula (Δg/g = ΔL/L + 2(Δt/t)), making it typically the dominant source of error. Check the error breakdown in the simulation results.
