Experimental Design With Period Measurements
- Identify which variables do and do not affect the period of a spring and a pendulum
- Design a controlled experiment to measure k or g from period data
- Linearize a period relationship so a slope yields the desired quantity
The two period formulas, and what is missing from them
For a mass on a spring, T = 2π√(m/k). For a simple pendulum at small angles, T = 2π√(L/g). What is absent matters as much as what is present. Neither contains amplitude — that independence is called isochronism and is what made pendulum clocks possible. And the pendulum formula contains no mass: a heavy bob and a light one on equal strings keep identical time, because gravity supplies both the restoring force and the inertia, and they cancel.
A spring oscillates the same on the Moon
The spring formula has no g in it, so a mass on a spring keeps the same period anywhere — on the Moon, in orbit, mounted horizontally or vertically. Hanging the spring vertically shifts the equilibrium position downward by mg/k but does not change the period, because the restoring force about that new equilibrium is still −kx with the same k. A pendulum, by contrast, is entirely dependent on g and would run about 2.4 times slower on the Moon.
Designing the experiment
A well-designed period experiment has three features the rubric looks for. Vary one variable across a wide range and hold the others fixed — five or six values of L, not two. Time many oscillations and divide, because timing 20 swings and dividing by 20 cuts your reaction-time error by a factor of 20. And plot to get a straight line, since a best-fit line through many points is far more trustworthy than averaging individual calculations of g.
Linearizing a square root
T = 2π√(L/g) plotted against L is a curve, and a curve has no usable slope. Square both sides: T² = (4π²/g)L. Now plot T² against L and you get a straight line through the origin with slope 4π²/g — so g = 4π²/slope. The same move works for the spring: T² against m gives slope 4π²/k. Whenever a period formula has a square root in it, squaring the period is the linearization the exam wants.
A student measures the period of a pendulum at several lengths and plots T² against L. The best-fit line has slope 4.1 s²/m. What is g at that location?
- 1.From T = 2π√(L/g), square both sides: T² = (4π²/g)L.
- 2.So the slope of a T²-vs-L graph is 4π²/g.
- 3.Rearrange: g = 4π² ÷ slope = 39.5 ÷ 4.1.
- 4.Evaluate: g ≈ 9.6 m/s².
The pendulum formula is a small-angle approximation, valid to about 15°. Beyond that the restoring force stops being proportional to displacement, the motion is no longer simple harmonic, and the measured period runs longer than the formula predicts. A design question that swings the bob to 60° is testing whether you notice.
A pendulum clock keeps correct time on Earth. Taken to the Moon, where g is about 1/6 as strong, it will:
To find the spring constant k from period data, which plot gives a straight line whose slope leads directly to k?
A student wants to reduce the effect of her reaction time when measuring a period with a stopwatch. The best method is to:
Experimental-design questions want a procedure, not just an equation: name the quantity you vary, the quantities you hold constant, the instrument you use, and what you plot. Each of those is typically its own rubric point, and they can be earned even if the final calculation goes wrong.
Answer the 3 checkpoints as you read.
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