Find the period of small oscillations in a vertical plane performed by a ball of mass m = 40 g fixed at the middle of a horizontally stretched string l = 1.0 m in length. The tension of the string is assumed to be constant and equal to F = 10 N.
The restoring force is 2F y/h = (4F/l) y, so the period of small oscillations is T = 2π√(ml/(4F)). With m = g, l = m and F = N, T = s.
Problem statement
Problem 4.18. Find the period of small oscillations in a vertical plane performed by a ball of mass m = 40 g fixed at the middle of a horizontally stretched string l = 1.0 m in length. The tension of the string is assumed to be constant and equal to F = 10 N.
Interpretation. The string is fixed at both ends, and the ball sits at its middle. The ball moves up and down in the vertical plane. Gravity only shifts the middle line a little, so it does not change the period, and the page leaves it out. The tension F stays the same as the ball moves.
We assume the swing is small, assumed 5 percent of the half-length of the string. The swing is a slider, so the page can also show a large swing, where the small-oscillation answer stops being true. The mass of the string is neglected.
Symbols: m is the mass of the ball, l the length of the string, h = l/2 the length of each half, F the tension, y the sideways shift of the ball, k the spring constant of the small-oscillation model and T0 = T the period.
The big idea in plain words
A ball pulled sideways on a tight string is pulled back by both halves of the string. For a small shift, the pull is nearly straight back and grows in proportion to the shift. That is exactly how a spring behaves. So find the spring constant, then use the spring formula.
The solution in each frame
The ball moves in the lab frame and nothing else moves, so only one frame is needed.
The solution step by step
Step 1, the force from one half. Shift the ball up by y. Each half of the string has length about h = l/2 and makes an angle θ with the horizontal, where tan θ = y/h. The tension F pulls along the string. Its vertical part is F sin θ, pointing down.
Step 2, small angle. For a small angle, sin θ is nearly tan θ = y/h. So each half gives a downward force F y/h. Two halves give 2F y/h = (4F/l) y. So Frestoring = −ky with k = 4F/l = N/m.
Step 3, the equation of motion. m y″ = −(4F/l) y, so ω0 = √(4F/(ml)) = s−1.
Step 4, the period. T = 2π/ω0 = 2π√(ml/(4F)). With m = 0.040 kg, l = 1.0 m and F = 10 N: T = 2π√(0.040 × 1.0/40) = 2π√0.001 = 0.199 s, about 0.2 s. On the page, T = s for the values set.
Step 5, the cases. The exact force is 2F y/√(y² + h²). For a large y this grows more slowly than y, so a big swing is slower than a small one. For the swing on screen the real period is s, a ratio of to T0. For a very large swing, a real string would also stretch, so the tension would not stay constant.
A separate calculation takes a tiny shift, computes the exact force from the angle of the two halves, and divides by the shift to get k. It also steps the exact equation of motion in small steps of time for a tiny swing and times a quarter swing. It compares both with the formulas. The largest relative difference is .
Five methods
1. Force method
Steps 1 to 4 above: the pull of the two halves, the spring form, the period.
2. Energy method
Stretching the string by moving the ball sideways stores energy 2F(√(y² + h²) − h), which is nearly F y²/h for small y. Compare with ½ky²: k = 2F/h = 4F/l. Same result.
3. Graph method
The first graph shows the displacement against the phase, with the grey dashes for the small-swing answer. The second shows how the real period grows with the swing. The third shows the stretch energy and the kinetic energy trading places while the total stays fixed.
4. Calculus method
The stored energy U(y) = 2F(√(y² + h²) − h) has U″(0) = 2F/h = 4F/l. Then ω0² = U″(0)/m.
5. Check method
Units: m l/F is in kg m/N = kg m/(kg m s−2) = s², so T is in seconds. Limits: a tighter string (larger F) or a shorter string (smaller l) makes a stiffer spring and a shorter period. A heavier ball makes a longer period.
Fastest: the force method. Most reliable: the numerical check.
What each slider changes
| Control | Changes the answer? | What it changes |
|---|---|---|
| Mass m | Yes | ω0, the period (it grows with the square root of m) and the speeds. It does not change the string |
| Length l | Yes | The half-length h, the restoring force, ω0, the period and the drawing |
| Tension F | Yes | The restoring force, ω0, the period and the energies |
| Start of the real motion | 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 m | The page finds m for a chosen period, and warns if it needs m outside 10 to 200 g |
| 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 |
|---|---|---|
| Mass of the ball | m | g |
| Length of the string | l | m |
| Tension | F | N |
| Start of the real motion | ε | % of h |
| Half-length | h | m |
| Spring constant | k | N/m |
| Period of small oscillations | T0 | s |
Exam traps
- Using the whole length l as the length of one half. Each half has length l/2, so the force is 2F y/(l/2), not 2F y/l.
- Counting only one half of the string. Both halves pull, so the force doubles.
- Putting the mass in grams into the formula. Use kilograms: 40 g = 0.040 kg.
- Adding gravity to the force. It only shifts the middle line and leaves the period unchanged.