Interactive physics simulator
Simple Pendulum
Explore gravity pendulum dynamics in three interactive modes. Compare the linear small-angle approximation against exact nonlinear orbits, inspect force vector balances and energy conservation, and analyze damped phase space spirals.
Simple Pendulum Dynamics Laboratory
Modify physical parameters in the right panel. Watch the pendulum swing and observe the live telemetry and curves update.
Live Pendulum Telemetry
- Elapsed Time
- 0.00 s
- Linear Angle (θ)
- 0.00°
- Nonlinear Angle (θ)
- 0.00°
- Linear Period (T0)
- 0.00 s
- Nonlinear Period (T)
- 0.00 s
Understanding the Simple Pendulum
A **Simple Pendulum** consists of a concentrated mass (known as the bob) attached to the end of a lightweight, inextensible string of length L suspended from a fixed pivot. When pulled back to an initial angle and released, the gravitational force exerts a restoring torque that drives it to swing back and forth, tracing a circular arc.
In physics, the simple pendulum serves as a key gateway to understanding rotational dynamics, periodic motion, and nonlinear oscillations.
The Restoring Force Equation
When displaced by an angle θ, the forces acting on the bob are Gravity (mg) downwards and Tension (T) along the string. Resolving gravity into components perpendicular and parallel to the string:
- **Radial component**: mg cosθ, which acts directly opposite to string tension.
- **Tangential component**: mg sinθ, which acts opposite to displacement, pulling the bob back to equilibrium.
Therefore, the tangential restoring force is:
The Small-Angle Approximation
Applying Newton\'s second law for rotational motion (τ = I · α) to the pendulum yields the exact nonlinear differential equation of motion:
Because this equation involves the sine of the angle (sinθ), it cannot be solved in terms of simple trigonometric functions. However, if the displacement angle is small (usually θ ≤ 15°), we can use the Taylor series approximation: **sinθ ≈ θ** (in radians).
Substituting this approximation linearizes the equation of motion:
This matches the standard equation for **Simple Harmonic Motion** with an angular frequency of ω = √(g/L). The period of oscillation is:
This elegant result demonstrates that, for small angles, the pendulum\'s period depends **only** on its length and the local gravity, remaining completely independent of both the mass of the bob and the amplitude of the swing.
Nonlinear Swings & Large-Angle Period
When the initial angle is large, the small-angle approximation fails because sinθ < θ. This means the restoring force is weaker than linear SHM assumes, causing the pendulum to move slower and take longer to complete a full swing.
The exact period T must be calculated using elliptic integrals, which can be expanded as an infinite series:
For example, at an initial angle of 45°, the actual period is about 4% longer than the small-angle period T_0. At 90°, it is 18% longer. In our simulator's first mode, you can observe this phase drift dynamically.
Energy Conservation in Pendulums
During a frictionless swing, total mechanical energy is conserved, continuously converting between Potential and Kinetic forms:
- Potential Energy (U): Stored as gravitational height. Taking the lowest point as y = 0, the height at angle θ is h = L(1 - cosθ). Thus:
U(θ) = m · g · L · (1 - cosθ)This peaks at the highest endpoints of the swing where the bob stops momentarily.
- Kinetic Energy (K): The energy of speed.
K = ½ m v²This peaks at the bottom crossing (equilibrium) where potential energy is zero.
Solved Examples
A simple pendulum has a string of length L = 1.0 meter. Find (a) the time period of its small-amplitude oscillations, and (b) the length of a "seconds pendulum" (a pendulum with a period of exactly 2.0 seconds) at the same location (g = 9.8 m/s²).
View Step-by-Step Solution
- Identify parameters: length L = 1.0 m, acceleration due to gravity g = 9.8 m/s².
- Part (a): Under small-angle approximation, use the period formula T = 2π√(L / g).
- T = 2 · 3.14159 · √(1.0 / 9.8) = 2 · 3.14159 · √(0.102) ≈ 6.283 · 0.319 ≈ 2.006 seconds.
- Part (b): For a seconds pendulum, T = 2.0 seconds. Rearrange the formula to solve for length: L = g · (T / 2π)².
- Substitute values: L = 9.8 · (2.0 / (2 · 3.14159))² = 9.8 · (1 / 3.14159)² = 9.8 · (0.3183)² = 9.8 · 0.1013 ≈ 0.993 meters.
- The time period is 2.01 s and the length of a seconds pendulum is 99.3 cm.
**Final Answer:** T = 2.01 s, L_seconds = 99.3 cm
A simple pendulum bob has a mass of 0.20 kg and is suspended by a 1.5-meter string. The pendulum is displaced to an angle of 30° and released from rest. Find (a) the tension in the string at the maximum displacement, and (b) the tension as the bob passes through the lowest point (equilibrium) at velocity v = 1.98 m/s.
View Step-by-Step Solution
- Identify parameters: mass m = 0.20 kg, length L = 1.5 m, gravity g = 9.8 m/s², max angle θ_0 = 30°.
- Part (a): At the maximum displacement, the velocity is zero. The radial forces are Tension (T) pulling inward and the radial component of gravity (mg cosθ_0) pulling outward.
- T = m · g · cos(θ_0) + m · (v² / L) = 0.20 · 9.8 · cos(30°) + 0 = 1.96 · 0.866 ≈ 1.70 Newtons.
- Part (b): At the lowest point, the angle θ = 0° (cosθ = 1) and the velocity v = 1.98 m/s.
- The tension must supply both the radial gravity balance and the required centripetal force: T = m · g + m · (v² / L).
- Substitute values: T = (0.20 · 9.8) + (0.20 · 1.98² / 1.5) = 1.96 + (0.20 · 3.92 / 1.5) = 1.96 + 0.52 = 2.48 Newtons.
- The tension is 1.70 N at maximum displacement and 2.48 N at the lowest point.
**Final Answer:** T_max_disp = 1.70 N, T_equilibrium = 2.48 N
Show that the small-angle approximation for a simple pendulum introduces less than a 2% error in the calculated period for initial displacement angles under 30°. Use the first correction term of the infinite series period formula.
View Step-by-Step Solution
- The exact period of a pendulum can be approximated by the series: T = T_0 [ 1 + (1/4) sin²(θ_0/2) + (9/64) sin&sup4;(θ_0/2) + ... ], where T_0 = 2π√(L/g) is the small-angle period.
- For θ_0 = 30°, calculate θ_0/2 = 15°. Find sin(15°) ≈ 0.2588.
- Calculate the first correction term: (1/4) · sin²(15°) = 0.25 · (0.2588)² = 0.25 · 0.0670 ≈ 0.0167.
- This means the actual period T is approximately T_0 · (1 + 0.0167) = 1.0167 · T_0.
- The error is (T - T_0) / T_0 ≈ 0.0167, which represents a 1.67% percentage error.
- Since 1.67% is less than 2%, the small-angle approximation is highly accurate for amplitudes up to 30°.
**Final Answer:** Error ≈ 1.67% (< 2%)
Common Misconceptions & Pitfalls
- Misconception: A heavier bob swings slower than a lighter bob.
**Reality:** Gravitational acceleration is independent of mass. The restoring force is proportional to mass (mg sinθ), but the inertia to be overcome is also proportional to mass. The mass cancels out completely in the acceleration equation, meaning both heavy and light bobs swing at the exact same rate. - Misconception: The formula T = 2π√(L/g) is exact for all angles.
**Reality:** This formula is derived using the linear approximation sinθ ≈ θ. It is accurate only for small angles. At large angles, the restoring force is weaker relative to displacement, increasing the period. - Misconception: The tension in the string is equal to mg at the lowest point.
**Reality:** At the lowest point, the string must support the bob\'s static weight (mg) **plus** provide the upward centripetal force required to pull it along a circular track (m v²/L). Thus, tension is always greater than mg at the bottom, equal to mg + mv²/L.
Practice Questions
Question 1
Why does the period of a simple pendulum remain independent of the mass of the bob?
Show Explanation
The restoring force on a pendulum bob displaced by an angle θ is the tangential component of gravity: F_t = -mg sinθ. According to Newton's second law, this force produces a tangential acceleration: a_t = F_t / m = -g sinθ. Notice that the mass (m) appears in both the force and the inertia terms and cancels out completely. This shows that all masses at a given location undergo the exact same gravitational acceleration, causing them to complete swings at the same rate. This is the same reason why all objects fall at the same rate in a vacuum.
Question 2
How does the period of a simple pendulum change if it is taken from Earth to the Moon, where gravity is one-sixth of Earth's gravity?
Show Explanation
The period of a simple pendulum is inversely proportional to the square root of the gravitational acceleration: T ∝ 1/√g. If gravity decreases on the Moon (g' = g / 6), the new period T' becomes: T' = 2π√(L / (g/6)) = √6 · [2π√(L / g)] = √6 · T. Since √6 ≈ 2.45, the period of the pendulum increases by approximately 2.45 times on the Moon, meaning the pendulum swings much slower.
Question 3
Why does the small-angle approximation break down at large initial angles, and what is its physical effect on the period?
Show Explanation
The small-angle approximation replaces sinθ with the angle θ in radians. This assumes a linear restoring force (F ∝ θ) which leads to perfect simple harmonic motion. However, as the displacement angle grows, sinθ becomes significantly smaller than θ (for example, at 90°, θ ≈ 1.57 rad while sinθ is only 1.0). Physically, this means the actual restoring force is weaker than the linear approximation predicts. A weaker restoring force causes the bob to accelerate slower, which increases the time needed to complete a full swing, making the actual period longer than T_0 = 2π√(L/g).
Question 4
Explain how tension in the pendulum string changes during a swing, and identify where the tension is minimum and maximum.
Show Explanation
The tension in the string is given by the formula T = mg cosθ + m v²/L. This formula shows that tension depends on the angle (gravity component) and the velocity (centripetal force component). At the highest points of the swing (θ = ±θ_0), the velocity is zero, so tension is at its minimum: T_min = mg cos(θ_0). As the pendulum swings down, the angle decreases (cosθ increases) and velocity increases. At the lowest point (θ = 0°), cosθ reaches its maximum value of 1, and the speed v reaches its peak. Thus, tension is at its absolute maximum at the bottom of the swing: T_max = mg + m v²/L.
Frequently Asked Questions
- What is a simple pendulum?
- A simple pendulum is an idealized physical model consisting of a point mass (bob) suspended by a massless, inextensible string from a frictionless pivot.
- What is the formula for the period of a simple pendulum?
- For small angles (less than 15°), the period is given by T = 2π√(L/g), where L is the string length and g is the acceleration due to gravity.
- Does the amplitude of oscillation affect the period?
- For small angles, the period is independent of amplitude. However, at large angles (above 20°), the period increases slightly due to the nonlinear restoring force.
- What forces act on a swinging pendulum bob?
- The two primary forces are gravity (pulling vertically downwards) and tension (pulling along the string towards the pivot point).
- What is a seconds pendulum?
- A seconds pendulum is a pendulum whose period is exactly 2.0 seconds (taking 1.0 second for a single swing from one side to the other). Its length is approximately 0.994 meters on Earth.
- What is the difference between a simple pendulum and a physical pendulum?
- A simple pendulum assumes all mass is concentrated at a single point (the bob) at the end of a massless string. A physical pendulum is a real rigid body of arbitrary shape that rotates about a pivot.