Aim
To investigate how the inertia of an object, shown by its period of oscillation on an inertial balance, depends on its mass.
Variables
- Manipulated: Mass loaded on the inertial balance, m
- Responding: Period of horizontal oscillation, T
- Constant: The inertial balance (its stiffness) and the amplitude of oscillation
Apparatus & materials
- Inertial balance (or a hacksaw blade clamped to the bench)
- Slotted masses
- G-clamp
- Stopwatch
- Retort stand
Procedure
- Clamp the inertial balance firmly to the edge of the bench so its pan can swing horizontally.
- Place a mass m = 100 g on the pan.
- Displace the pan a fixed distance to one side and release it so it oscillates horizontally.
- Time 10 complete oscillations and record t; calculate T = t / 10.
- Repeat for m = 200, 300, 400 and 500 g.
- Record m, t and T, and calculate T² for each mass.
Tabulating results
Record mass m in g (or kg), the time for 10 oscillations t in s, the period T = t / 10 in s, and T² in s². Keep the decimal places consistent.
The graph
Plot T² (y-axis) against m (x-axis). The straight line shows that a larger mass gives a longer period, and T² is directly proportional to m.
Analysis
A greater mass has greater inertia, so it resists the change in its motion more and takes longer to complete each oscillation. The period T (and T²) therefore increases as the mass increases.
Precautions
- Use the same amplitude each time so only the mass is changed.
- Make sure the clamp holds the balance firmly so no energy is lost to the bench.
- Count the oscillations from the same point of the swing.
Force and Motion I · Graph skills
Sample results and what they show
These are example readings, not a mark scheme. For masses m = 100, 200, 300, 400, 500 g on the inertial balance, the period T rose from about 0.40 s to 0.89 s, giving T² of about 0.16, 0.32, 0.48, 0.64 and 0.80 s².
The pattern is that a heavier load swings back and forth more slowly. The mass has more inertia, so it resists the blade's changing motion more strongly and each oscillation takes longer.
Because the times are short, count 10 (or more) oscillations and divide, rather than timing one. The derived column shows the real relationship: T² is directly proportional to m, T² doubles from 0.16 s² to 0.32 s² as m doubles from 100 g to 200 g.
This is a horizontal oscillation, so gravity is not the restoring force; the springiness of the blade is, which is why the result depends on inertia rather than weight.
Reading the graph and finding the answer
Plot T² / s² on the y-axis against m / kg on the x-axis. The points give a straight line, showing T² increases in direct proportion to the mass, greater mass, greater inertia, longer period.
Take the gradient from a large triangle on the best-fit line. Using (0.100 kg, 0.16 s²) and (0.500 kg, 0.80 s²): gradient = (0.80 − 0.16) s² ÷ (0.500 − 0.100) kg = 0.64 s² ÷ 0.400 kg = 1.6 s² kg⁻¹.
The steady positive gradient is the quantitative statement that inertia rises with mass; for the inertial balance T² = (4π²/k) m, so the gradient equals 4π²/k, where k is the stiffness of the blade. The key reading of the graph is simply that a straight line of positive gradient links period-squared to mass, confirming the aim.
Marks examiners look for
State the variables: manipulated m / g, responding T / s, with the same balance (its stiffness) and the same amplitude kept constant. Using the same starting displacement every time is the fair-test point, since the period must depend on mass alone, not on how hard you push the pan.
Clamp the blade firmly so no energy leaks into the bench, and count oscillations from the same point of the swing. Tabulate m / g, t / s for 10 oscillations, T / s and T² / s², with units in the headings and consistent decimals.
On the graph, draw a single best-fit straight line and comment that its positive gradient shows inertia increasing with mass. A conclusion that states T² ∝ m and explains it in terms of inertia, a larger mass resisting the change in motion more, secures the analysis mark.
Avoid describing this as a gravity or weighing experiment; on a horizontal balance the result is set by inertia.
Source: DSKP KSSM Physics Form 4 and 5 (Versi English) (Bahagian Pembangunan Kurikulum (BPK), KPM)