Question 1 — Spring oscillations: series vs parallel stiffness
Time oscillations of a mass on a 'double spring' (two springs in series) and then on two springs in parallel, then vary the mass and plot √T2 against √T1.
RUN IT ON THE BENCH
Your readings feed straight into the answer table below as you record them.
ON YOUR BENCH — FROM THE CONFIDENTIAL INSTRUCTIONS
Question 1 — apparatus
- stand(2)
- boss(2)
- expendable springs(4)
- longer wooden rod(1)
- shorter wooden rod(1)
- 100 g mass hanger(1)
- 10 g slotted mass(2)
- 50 g slotted mass(1)
- 100 g slotted mass(2)
- stopwatch(1)
- 180° protractor(1)
- metre rule(1)
THE BENCH — MASS HANGING FROM TWO SPRINGS
Pull the mass down a short, consistent distance and release — don't push it sideways.
Why is this a different experiment from the pendulum, when both time oscillations?
A pendulum's restoring force is gravity acting along a swinging arc, so its period depends on length: T = 2π√(L/g). Here the restoring force is a spring obeying Hooke's law, F = −kx, so the period depends on stiffness and mass instead: T = 2π√(m/k). Same shape of formula, completely different physics underneath it — which is exactly why this needs its own apparatus rather than reusing the pendulum's string and bob.
Why does T2 come out about half of T1?
Two identical springs in series behave like one long, floppy spring — half the stiffness of one alone, so k_series = 12.5 N/m. The same two springs in parallel share the load, so they add: k_parallel = 50 N/m, four times stiffer than the series case. Since T ∝ 1/√k, quadrupling k halves the period — T2 should come out close to half of T1 for any mass you choose, not just the 270 g starting value.
Why plot √T2 against √T1, rather than T2 against T1?
T1 = 2π√(m/k_series) and T2 = 2π√(m/k_parallel), so both √T1 and √T2 are proportional to m1/4 — meaning √T2 is directly proportional to √T1 itself, giving a straight line through the origin instead of a curve. Since T2 = 0.5·T1 at every mass (because T2/T1 = √(k_series/k_parallel) = √0.25 = 0.5), it follows that √T2 = √(0.5·T1) = √0.5·√T1 — so the gradient of √T2 against √T1 is √0.5 ≈ 0.707, not 0.5 itself. Check that against the gradient your own five points give below.
YOUR TASKS · 0/3
- ✓Time T1 (series) and T2 (parallel) at the starting mass, 270 g
- ✓Repeat for four more masses, none below 200 g — five sets in total
- ✓Read off the gradient of √T2 against √T1
TABLE 1(c) — YOUR FIVE SETS
Time T1 and T2 at 270 g, then save your first set.
√T2 vs √T1
√T2 = P·√T1 + Q. Since T2/T1 = √(k_series/k_parallel) = √0.25 = 0.5 at every mass, theory predicts P = √(T2/T1) = √0.5 ≈ 0.707 and Q ≈ 0 — a straight line through the origin, a little less steep than y = x.
YOUR ANSWERS
Slide the double spring and the two single springs onto the longer wooden rod (Fig. 1.1). Fix the rod approximately 55 cm above the bench. Hang a total mass of 270 g from the double spring; this mass is m. Record m.
Gently pull the mass down through a short distance and release. Take measurements to determine the period T1 of the oscillations.
Using the shorter wooden rod (with its two notches), hang mass m from the two single springs in parallel instead (Fig. 1.2), keeping the rod level. Pull down gently and release, then take measurements to determine the period T2 of the oscillations.
Vary m and, for each value, determine T1 and T2 as above. Do not use values of m less than 200 g. Repeat until you have five sets of values of m, T1 and T2. Record your results in a table, including values of √T1 and √T2.
| Set 1 | Set 2 | Set 3 | Set 4 | Set 5 | |
|---|---|---|---|---|---|
| m / g | |||||
| T1 / s | |||||
| √T1 / s0.5 | |||||
| T2 / s | |||||
| √T2 / s0.5 |
Plot a graph of √T2 (y-axis) against √T1 (x-axis) using your results from 1(c).
Draw the straight line of best fit through your plotted points.
Determine the gradient and the y-intercept of your line of best fit.
It is suggested that √T2 = P√T1 + Q, where P and Q are constants. Using your answers to 1(d)(iii), determine the values of P and Q, giving appropriate units.