Derived by combining F = -kx (Hooke's law) with F = ma and the SHM condition a = -$\omega$^2 x
Oscillations - OCR A-Level Physics
$$T = 2\pi\sqrt{\frac{m}{k}}$$
- Derived by combining F = -kx (Hooke's lawThe extension of a spring is directly proportional to the applied force, provided the limit of proportionality is not exceeded.) with F = ma and the SHM condition a = -$\omega$^2 x.
- This $gives \omega$^2 = k/m, $so \omega =$ $\sqrt{k/m}$and T = $2\pi/\omega$ = $2\pi\sqrt{m/k}.$
- The periodThe time taken for one complete oscillation or wave cycle. Measured in seconds (s). does NOT depend on amplitudeThe maximum displacement of a point on a wave from its equilibrium (rest) position. Measured in metres (m). (isochronous).
- Increasing mass increases the periodThe time taken for one complete oscillation or wave cycle. Measured in seconds (s). (heavier objects oscillate more slowly).
- Increasing the spring constantThe force per unit extension of a spring. A measure of the stiffness of the spring. Measured in N m⁻¹. decreases the periodThe time taken for one complete oscillation or wave cycle. Measured in seconds (s). (stiffer springs oscillate faster).
- For springs in parallelComponents connected across the same two points, providing multiple current paths.: $k_{eff}$ = k_1 + k_2. For springs in seriesComponents connected end-to-end in a single path, so current flows through each in turn.: 1/$k_{eff}$ = 1/k_1 + 1/k_2.
Worked Example
A 0.40 kg mass oscillates on a spring with k = 160 $N m^{-1}$. Calculate the period and frequencyThe number of complete oscillations passing a point per unit time. Measured in hertz (Hz)..
Show Solution
1
$T = 2\pi\sqrt{m/k} = 2\pi\sqrt{0.40/160}.$
2
$T = 2\pi\sqrt{2.5 \times 10^{-3}}.$
3
$T = 2\pi \times 0.0500 = 0.314 s.$
4
$f = 1/T = 1/0.314 = 3.18 Hz.$
Answer
$T = 0.31 s, f = 3.2 Hz$