Frequency and Amplitude of a Spring Oscillator

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Oscillations and Waves Beginner Simple harmonic oscillation

Source: High school physics (Chinese)

Problem

A 5.0 kg object moves on a frictionless horizontal surface under a spring of force constant $1.0 \times 10^3$ N/m. The object is displaced to 50 cm from the equilibrium position and given an initial speed of 10 m/s directed toward the equilibrium position.

  1. Find the frequency of the oscillation.
  2. Find the amplitude of the oscillation.

(1) $f = \frac{1}{2\pi}\sqrt{k/m} \approx 2.3$ Hz. (2) $A = \sqrt{0.75} \approx 0.87$ m.

(1) The angular frequency depends only on $k$ and $m$:

$$\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{1.0\times 10^3}{5.0}} = \sqrt{200} \approx 14.1 \text{ rad/s},$$

so the frequency is

$$f = \frac{\omega}{2\pi} = \frac{\sqrt{200}}{2\pi} \approx 2.3 \text{ Hz}.$$

(2) By energy conservation the amplitude follows from the displacement $x_0 = 0.50$ m and speed $v_0 = 10$ m/s:

$$A = \sqrt{x_0^2 + \frac{v_0^2}{\omega^2}} = \sqrt{(0.50)^2 + \frac{(10)^2}{200}} = \sqrt{0.25 + 0.50} = \sqrt{0.75} \approx 0.87 \text{ m}.$$