Valid only for small angles of oscillation (less than about $10\degree$) $where \sin\theta$
Oscillations - OCR A-Level Physics
$$T = 2\pi\sqrt{\frac{l}{g}}$$
- Valid only for small anglesThe approximation sin θ ≈ θ (in radians) used when deriving the time period of a simple pendulum; valid for θ < ~10°. of oscillation (less than about $10\degree$) $where \sin\theta$ $\approx \theta.$
- The periodThe time taken for one complete oscillation or wave cycle. Measured in seconds (s). does NOT depend on mass or amplitudeThe maximum displacement of a point on a wave from its equilibrium (rest) position. Measured in metres (m). (for small angles).
- Increasing length increases the periodThe time taken for one complete oscillation or wave cycle. Measured in seconds (s)..
- A pendulum on the Moon (lower g) would have a longer periodThe time taken for one complete oscillation or wave cycle. Measured in seconds (s). than on Earth.
- This equation can be used to measure g experimentally: g = $4\pi$^2 l / \(T^{2}\). Plot \(T^{2}\) vs l and the gradient = $4\pi$^2/g.
Worked Example
A pendulum of length 1.2 m oscillates with small amplitudeThe maximum displacement of a point on a wave from its equilibrium (rest) position. Measured in metres (m).. Calculate its period on Earth (g = 9.81 $m s^{-2}$).
Show Solution
1
$T = 2\pi\sqrt{l/g} = 2\pi\sqrt{1.2/9.81}.$
2
$T = 2\pi\sqrt{0.1223}.$
3
$T = 2\pi \times 0.3497 = 2.198 s.$
4
$$T \approx 2.2$ s.$
Answer
$$T \approx 2.2$ s$
Common Mistake
MEDIUM
Wrong: Measuring the length of a pendulum to the top of the bob.
Right: The length l is measured from the pivot to the CENTRE OF MASS of the bob, not the top or bottom.
Right: The length l is measured from the pivot to the CENTRE OF MASS of the bob, not the top or bottom.