Determine the period of small oscillations of a mathematical pendulum, that is a ball suspended by a thread l = 20 cm in length, if it is located in a liquid whose density is η = 3.0 times less than that of the ball. The resistance of the liquid is to be neglected.
In the liquid, gravity acts as g′ = g(1 − 1/η), so T = 2π√(l/g′) = 2π√(ηl/(g(η − 1))). With l = m and η = , g′ = m/s² and T = s.
Problem statement
Problem 4.19. Determine the period of small oscillations of a mathematical pendulum, that is a ball suspended by a thread l = 20 cm in length, if it is located in a liquid whose density is η = 3.0 times less than that of the ball. The resistance of the liquid is to be neglected.
Interpretation. The ball hangs from a thread and swings in a small arc, all of it inside the liquid. The liquid pushes the ball up (buoyancy) but does not slow it down, because the resistance is neglected. The thread has no mass and does not stretch.
We assume g = 9.8 m/s², and the swing is small, assumed 5 degrees. The swing is a slider, so the page can also show a large swing, where the small-angle answer stops being true. The liquid that moves with the ball (the added mass) is also neglected, as the problem says to neglect the effects of the liquid other than buoyancy.
Symbols: l is the length of the thread, η the ratio of the ball density to the liquid density, g′ the effective gravity in the liquid, ω0 the angular frequency and T0 = T the period.
The big idea in plain words
In air, a pendulum is pulled by the weight mg. In the liquid, part of the weight is cancelled by the push of the liquid. What is left behaves like a smaller weight mg′. So the pendulum swings like an ordinary pendulum in a world with weaker gravity g′. Weaker gravity means a longer period.
The solution in each frame
The ball swings in the lab frame, and the liquid is at rest in it, so only one frame is needed.
The solution step by step
Step 1, the buoyancy. The liquid pushes up with a force equal to the weight of the liquid the ball displaces. If the ball has volume V and density ρ, its mass is m = ρV. The liquid has density ρ/η, so the push is (ρ/η) V g = mg/η.
Step 2, the effective weight. The net downward force is mg − mg/η = mg(1 − 1/η). Write this as mg′, so g′ = g(1 − 1/η) = m/s². The buoyancy always points straight up, just like the weight, so it only changes the strength of gravity.
Step 3, the pendulum. For a small angle θ, the sideways force is mg′ sin θ, nearly mg′θ. The equation of motion is l θ″ = −g′ θ, so ω0 = √(g′/l) = s−1.
Step 4, the period. T = 2π√(l/g′) = 2π√(ηl/(g(η − 1))). With l = 0.20 m and η = 3.0: T = 2π√(3.0 × 0.20/(9.8 × 2.0)) = 2π√0.0306 = 1.1 s. On the page, T = s for the values set.
Step 5, the cases. In air (η very large) g′ tends to g and T tends to 2π√(l/g). If η tends to 1, g′ tends to zero and the period grows without limit. For a swing of degrees the real period is s, a ratio of to T0.
A separate calculation takes a ball volume and the two densities, finds the weight and the buoyancy as forces, and gets g′ from the net force divided by the mass. It also steps the full equation of motion, with the sine and no small-angle rule, for a tiny swing and compares its period with the formula. The largest relative difference is .
Five methods
1. Force method
Steps 1 to 4 above: weight minus buoyancy, the effective gravity, the pendulum period.
2. Energy method
Raising the ball by a height h costs energy mg′h, because the liquid helps to lift it. For a small angle, h = lθ²/2, so U = ½(mg′/l)(lθ)². This is the energy of a spring with k = mg′/l. So ω0² = k/m = g′/l.
3. Graph method
The first graph shows the angle against the phase, with grey dashes for the small-swing answer. The second shows how the real period grows with the starting angle. The third shows the potential and the kinetic energy trading places.
4. Calculus method
The potential energy is U(θ) = mg′l(1 − cos θ). Its second derivative at the bottom is mg′l, and the moment of inertia is ml². So ω0² = mg′l/(ml²) = g′/l.
5. Check method
Units: l/g′ is in m/(m s−2) = s², so T is in seconds. Limits: η → ∞ gives the ordinary pendulum in air. η = 1 gives g′ = 0, no restoring force, and an infinite period. η below 1 would make the ball float, and the pendulum would hang upwards.
Fastest: the effective gravity idea. Most reliable: the numerical check.
What each slider changes
| Control | Changes the answer? | What it changes |
|---|---|---|
| Length l | Yes | ω0, the period (it grows with the square root of l) and the drawing |
| Density ratio η | Yes | The buoyancy arrow, g′, ω0 and the period |
| Starting angle | No | How far the real ball swings, the real period and the gap to the grey ghost. It does not change T0 |
| Phase ω0t | No | Where the ball and the dots are |
| Given-answer mode | Sets l | The page finds l for a chosen period, and warns if it needs l outside 0.05 to 1 m |
| Play and time bar | No | Play runs two small-oscillation periods in about 16 s and holds 3 s after one period. The bar sets the phase |
Parameters
| Quantity | Symbol | Value used |
|---|---|---|
| Length of the thread | l | m |
| Density ratio | η | |
| Starting angle | θ0 | degrees |
| Effective gravity | g′ | m/s² |
| Period of small oscillations | T0 | s |
Exam traps
- Writing g′ = g/η. The correct form is g(1 − 1/η): the buoyancy is subtracted from the weight.
- Using η the wrong way round. Here η is ball density divided by liquid density, and the liquid is lighter than the ball.
- Using l in centimetres. Put l = 0.20 m into the formula.
- Using the answer for a large swing. The period grows when the angle is large.