SME
Grade 11 Science · Problem type
A simple pendulum
Use this for “How long is a pendulum that swings once every 2 s? What would its period be on the Moon?”
Diagram
T
=
2π
√
L/g
=
2π
×
√
0.8/9.8
T = 1.795 s
f = 1/T = 0.557 Hz
For small swings T depends only on L and g: not on the bob’s mass or how far it swings.
This is an example. Type your own values to start a new problem.
T (s)
=
2π
L (m)
g (m/s²)
f
Frequency
calculated · Hz
Formulas
T
=
2π
×
√
L
g
1.795
=
2π
×
√
0.8/9.8
f
=
1
T
0.557 = 1/1.795
Assumptions
•
Small swings, under about 15°: then the period doesn’t depend on how far it swings.
•
The bob’s mass doesn’t change the period; only the length and gravity do.
•
g is 9.8 m/s² on Earth and 1.62 m/s² on the Moon.
Step-by-step
WE KNOW
L = 0.8 m, g = 9.8 m/s²
FIND
period (T), frequency (f)
Step 1 · Find period (T)
Use T = 2π × √(L/g)
A longer string swings more slowly; stronger gravity, faster.
T
=
2π
×
√
0.8/9.8
√
0.08163
=
0.2857
T
=
2π
×
0.2857
T
=
1.795
s
Step 2 · Find frequency (f)
Use f = 1/T
The frequency is how many swings fit in one second.
f
=
1/1.795
f
=
0.557
Hz
Check
Put the numbers back in. Both sides should match.
1.795
≈
2π
×
√
0.8/9.8
✓
0.557 ≈ 1/1.795 ✓
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