A ball is suspended by a thread of length l at the point O on the wall, forming a small angle α with the vertical (Fig. 4.1). Then the thread with the ball was deviated through a small angle β (β > α) and set free. Assuming the collision of the ball against the wall to be perfectly elastic, find the oscillation period of such a pendulum.
T = 2√(l/g) (π/2 + arcsin(α/β)). With l = m, α = degrees and β = degrees, T = s.
Problem statement
Problem 4.20. A ball is suspended by a thread of length l at the point O on the wall, forming a small angle α with the vertical (Fig. 4.1). Then the thread with the ball was deviated through a small angle β (β > α) and set free. Assuming the collision of the ball against the wall to be perfectly elastic, find the oscillation period of such a pendulum.
Interpretation. The wall is a plane through the point O, tilted by the small angle α from the vertical. The thread hangs along the wall when the ball rests against it. The ball is pulled to the other side of the vertical by the angle β and released from rest. It swings through the lowest point, hits the wall, bounces back and returns to the release point. One full trip there and back is one period.
We assume l = 1.0 m and g = 9.8 m/s², because the problem gives only symbols for the length. Both are controls. The angles are assumed α = 3 degrees and β = 6 degrees. The page also allows large angles, to show where the small-angle answer stops being true.
Symbols: l is the length of the thread, α the angle of the wall, β the release angle, ω = √(g/l), t1 the time from the release to the wall and T the period.
The big idea in plain words
Away from the wall, the ball is an ordinary pendulum, so its angle follows a cosine. The wall cuts the swing short, and an elastic bounce only turns the ball round. So the period is just twice the time the ball needs to get from the release point to the wall.
The solution in each frame
The ball and the wall are both at rest in the lab frame, so only one frame is needed.
The solution step by step
Step 1, the free motion. For small angles the ball obeys the pendulum equation l θ″ = −gθ. Measure θ from the vertical, positive on the release side. Released from rest at θ = β, the motion is θ = β cos ωt, with ω = √(g/l) = s−1.
Step 2, the time to the wall. The wall is at θ = −α. The ball reaches it when β cos ωt1 = −α. So cos ωt1 = −α/β. The first angle with this cosine is ωt1 = π/2 + arcsin(α/β) = rad. So t1 = s.
Step 3, the bounce. The hit is perfectly elastic, so the speed is unchanged and only its direction reverses. The ball moves back along the same path. It reaches β after another time t1, at rest, like at the start.
Step 4, the period. T = 2t1 = 2√(l/g)(π/2 + arcsin(α/β)) = s.
Step 5, the cases. If α = 0, the wall is vertical and arcsin 0 = 0, so T = π√(l/g), half of the free period. If α = β, arcsin 1 = π/2 and T = 2π√(l/g) = s, the free period, because the ball only touches the wall at the end of its swing. If β is less than α, the ball never reaches the wall. The exact motion for the angles set has period s, a ratio of to the formula.
A separate calculation steps the small-angle equation of motion in very small time steps, turns the velocity round when the ball reaches the wall, and times the return to the release point. It compares this time with the formula. The relative difference is . The page also steps the exact equation, with the sine and no small-angle rule, to show the effect of large angles.
Five methods
1. Force method
Steps 1 to 4 above: the pendulum equation, the time to the wall, the bounce, the period.
2. Energy method
The ball starts with energy mg l(1 − cos β), nearly ½ mg l β². An elastic hit loses no energy, so the swing always reaches β again. This confirms that the ball returns to the release point after the bounce.
3. Graph method
The first graph shows the angle against the phase. It is a cosine that is cut off at the wall level (the amber line) and turned back. The second graph shows how the period grows with the wall angle. The third shows the potential and the kinetic energy.
4. Calculus method
Solve θ″ = −ω²θ with θ(0) = β and θ′(0) = 0. Set θ = −α to find t1, then double it.
5. Check method
Units: √(l/g) is in √(m/(m s−2)) = s, and the bracket is a pure number. Limits: α → 0 gives half the free period, and α → β gives the free period. Both are what we expect.
Fastest: the cosine method. Most reliable: the numerical check.
What each slider changes
| Control | Changes the answer? | What it changes |
|---|---|---|
| Length l | Yes | ω, the period (it grows with the square root of l), the speeds and the drawing |
| Wall angle α | Yes | Where the ball hits the wall, t1, the period and the drawing |
| Release angle β | Yes | The swing, t1, the period and the drawing. If β is not larger than α, the ball never hits the wall |
| Phase | 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 0.1 to 2 m |
| Play and time bar | No | Play runs two 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 |
| Wall angle | α | degrees |
| Release angle | β | degrees |
| Time to the wall | t1 | s |
| Period | T | s |
| Free period | T0 | s |
Exam traps
- Taking the period as half of the free period for any wall angle. That holds only for α = 0.
- Using arccos(α/β) instead of π/2 + arcsin(α/β). The ball must pass the vertical and reach the wall on the far side.
- Forgetting to double t1. The period is the trip to the wall and back.
- Measuring the angles from the horizontal. Both α and β are measured from the vertical.