Conservation of Energy: Simple Pendulum

Pull the pendulum aside and let go — watch “height” energy turn into motion energy and back. The sum stays the same all the time — that is the law of conservation of energy. Switch on resistance and see where the energy goes.

The idea in 30 seconds

Lift it — store it

The higher the bob is lifted, the more energy is stored in it. This is potential energy — “energy of position”: Ep = mgh, where h is the height above the lowest point.

Release it — convert it

The bob falls: the height decreases and the speed grows. “Height energy” turns into motion energy — kinetic energy: Ek = mv²/2. At the lowest point it is the greatest.

The sum stays the same

Energy neither disappears nor appears from nowhere — it only changes from one kind into another: Ep + Ek = const. With resistance part of it goes into heat Q, but together with the heat the sum is the same.

Try it yourself

Start the pendulum and watch the column on the right: the orange and green parts “pour” into each other while the total height of the column stays the same.
Press “m ×2”: all the energies doubled, but the speed at the lowest point stayed the same. Why? The mass cancels in the formula v = √(2gh).
Press “h ×2” — the bob is lifted twice as high. The speed at the bottom grew, but not twice — only √2 ≈ 1.41 times.
Choose “Air” or “Water”: the red segment (heat Q) grows and the bob rises lower and lower below the yellow line h₀. Check that Ep + Ek + Q is still equal to E₀.

Where you meet it

A swing: at the highest point you stop for a moment (Ek = 0), at the bottom you fly the fastest (Ep = 0). To swing higher you have to add energy — push in time with the motion.
A pendulum clock: because of friction at the pivot the pendulum would stop, so a weight or a spring adds the lost energy on every swing.
A roller coaster and a ski jump: the speed at the bottom depends only on the height of the drop, v = √(2gh), and not on the mass (ignoring air resistance).

Graphs: what changes and how

The first graph is the “potential well”: the curve Ep(θ), the yellow line is the total energy E, the green segment is Ek. The bob always sits on the curve and turns back where it meets the yellow line. The other graphs are drawn live; the grey dashed line on the angle graph is the small-oscillation approximation: for a large swing the real pendulum lags behind it because the period depends on the amplitude.

The potential well: why the pendulum turns back

Energy over time: Ep and Ek flow into each other, the sum is flat

Angle over time θ(t) compared with small oscillations

Speed over time v(t): peaks at the lowest point

θ₀ release angle 60°

l string length 2,0 m

m bob mass 2,0 kg

Medium

b resistance of the medium 0,00 1/s

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gravity and speed

height levels h and h₀

motion trail

height ruler

“Predict the speed” mode

The string is weightless and inextensible, the bob is a point mass. The resistance of the medium is described by the coefficient b (1/s): the larger it is, the faster the pendulum loses energy. The lost mechanical energy turns into heat Q. The release angle is at most 90°: above the horizontal the string would go slack.

KEY FORMULAS

Height of the bob above the lowest point

$$h = l\,(1-\cos\theta)$$

The height is measured from the lowest point of the arc: there h = 0 and Ep = 0.

▶ Detailed Explanation

A string of length l is deflected by an angle θ from the vertical. The lowest point is a distance l below the pivot, while the bob itself is at l·cosθ. So the bob is raised above the lowest point by h = l − l·cosθ = l(1 − cosθ).

Example: l = 2 m, θ₀ = 60°. cos 60° = 0.5, so h₀ = 2·(1 − 0.5) = 1.00 m. Extreme cases: at θ = 0 the height is h = 0 (the lowest point); at θ = 90° (the string is horizontal) h = l — the bob cannot rise higher, which is why the simulator’s limit is 90°.

Conservation of energy for a pendulum

$$E_{\text{p}} + E_{\text{k}} = mgh + \frac{mv^2}{2} = E_0 = \text{const}$$

Without resistance the sum of potential and kinetic energy is the same at every point of the arc.

▶ Detailed Explanation

Ep = mgh is the energy of height, Ek = mv²/2 is the energy of motion. The string tension does no work (it is always perpendicular to the velocity), and gravity is “restoring”: what the bob loses in height it gains in speed. So at every point Ep + Ek = E₀ = mgh₀, where h₀ is the height it was released from.

Example: m = 2 kg, h₀ = 1.00 m ⇒ E₀ = 2·9.8·1.00 = 19.6 J. At a height h = 0.40 m: Ep = 2·9.8·0.40 = 7.84 J, so Ek = 19.6 − 7.84 = 11.76 J and the speed is v = √(2Ek/m) = √11.76 ≈ 3.43 m/s.

Speed at the lowest point

$$v = \sqrt{2gh_0} = \sqrt{2gl\,(1-\cos\theta_0)}$$

From mgh₀ = mv²/2: the mass cancels, the speed depends only on the height of release.

▶ Detailed Explanation

At the lowest point all the energy is kinetic: mgh₀ = mv²/2. Cancelling m gives v = √(2gh₀). A heavy and a light pendulum released from the same height pass the bottom with the same speed.

Example: l = 2 m, θ₀ = 60° ⇒ h₀ = 1.00 m, v = √(2·9.8·1.00) = 4.43 m/s. If the bob is lifted twice as high (h₀ = 2 m; for that you need l = 2 m and θ₀ = 90°), the speed grows only by √2 ≈ 1.41 times — to 6.26 m/s. The speed grows more slowly than the height. The limiting case θ₀ = 90°: v = √(2gl).

Energy with resistance: where the rest goes

$$E_{\text{p}} + E_{\text{k}} + Q = E_0,\qquad Q = E_0 - (E_{\text{p}} + E_{\text{k}})$$

Resistance turns part of the mechanical energy into heat, but the total energy including the heat does not change.

▶ Detailed Explanation

Because of air resistance and friction at the pivot each oscillation is a little lower than the previous one. The mechanical energy Ep + Ek decreases but does not vanish: it turns into heat Q (the air and the pivot warm up). In the simulator Q is computed from the equations of motion as the work of the resistance force — it is the red segment of the column.

Example: E₀ = 19.6 J. After a few swings Ep + Ek = 14.2 J — so Q = 19.6 − 14.2 = 5.4 J went into heat. When the pendulum stops, Q = E₀ = 19.6 J and the balance E₀ − Ep − Ek − Q stays zero.

Pendulum period and the swing

$$T_0 = 2\pi\sqrt{\frac{l}{g}},\qquad T \approx T_0\left(1+\frac{\theta_0^{2}}{16}\right)$$

The period depends on the string length and, for a large swing, slightly on the angle too (θ₀ in radians).

▶ Detailed Explanation

T₀ = 2π√(l/g) is the period for small angles: it depends neither on the mass nor on the swing. But a real pendulum is nonlinear (the angular acceleration is proportional to sin θ, not θ), and for a large swing the oscillations are a little slower. The correction: T ≈ T₀(1 + θ₀²/16), where θ₀ is in radians.

Example: l = 2 m ⇒ T₀ = 2π√(2/9.8) ≈ 2.84 s. At θ₀ = 60° (1.047 rad) the exact period is 3.05 s — 7 % longer; at θ₀ = 90° it is 3.35 s, 18 % longer. The grey dashed line on the angle graph is the oscillation with the period T₀: for a large swing the real curve lags behind it.

What is going on. A simple pendulum is a bob on a thin inextensible string. When we pull it aside, the bob rises to a height h above the lowest point of the arc, and we “store” energy of position in it — potential energy: Ep = mgh. If the bob is released with no push, at that moment all of its energy is potential and its speed is zero.

Conversion and conservation. The bob goes down: the height decreases, the speed grows, and potential energy turns into kinetic energy Ek = mv²/2. At the lowest point h = 0, Ep = 0 and all the energy is kinetic — so the speed is greatest there: v = √(2gh₀). The bob then swings up the other side by inertia — again up to the height h₀ — and everything repeats. As long as there is no resistance, the sum Ep + Ek is the same at every point: this is the law of conservation of mechanical energy. Example: m = 2 kg, l = 2 m, θ₀ = 60° ⇒ h₀ = 1.00 m, E₀ = 19.6 J, the speed at the bottom v = 4.43 m/s — and it does not depend on the mass.

Resistance and heat. Switch on the “Air” or “Water” medium: because of air resistance and friction at the pivot each swing is a little lower than the previous one. The mechanical energy Ep + Ek decreases but does not vanish — it turns into heat Q (the red segment in the column). Counting the heat, the sum is the same: Ep + Ek + Q = E₀.

Why not more than 90°, and what about the period. If the bob were released above the horizontal, the string would go slack (its tension T = mg·cosθ would be negative), so the release angle is limited to 90°. The pendulum equation is nonlinear: the angular acceleration is proportional to sin θ, not θ. For small angles the period T₀ = 2π√(l/g) depends neither on the mass nor on the swing; for a large swing the pendulum is a little slower (7 % longer period at 60°, 18 % at 90°) — you can see this in the grey dashed line on the angle graph. The simulator computes the exact nonlinear motion, not an approximation.

Compare it with the simulators “Spring pendulum: simple harmonic motion” and “Conservation of energy: roller coaster”: the period of a spring pendulum strictly does not depend on the amplitude, and on the roller coaster the bob slides down an arbitrary track — but the law of conservation of energy works in exactly the same way.