Question

Write the equation of motion of a simple harmonic oscillator which has an amplitude of
5 cm and it executes 150 oscillations in 5 minutes with an initial phase of 45°. Also
obtain the value of its maximum velocity.

Answer 1

Write the equation of motion of a simple harmonic oscillator which has an amplitude of

5 cm and it executes 150 oscillations in 5 minutes with an initial phase of $45{}^{\circ}$. Also obtain the value of its maximum velocity.

Solution:

$A = 5\,\mathrm{cm} = 0.05\,\mathrm{m}$ — amplitude of harmonic oscillations;

$\varphi_0 = 45{}^{\circ} = \frac{\pi}{4}$ — initial phase;

Equation of motion of a simple harmonic oscillator:

$$x(t) = A \cos(\omega t + \varphi_0)$$

To find the equation to find the cyclic frequency. Cyclic frequency shows how many radians moved the body per unit of time:

$$\omega = 2\pi \cdot f = 2\pi \cdot \frac{150\,\text{oscillations}}{5 \cdot 60\,\text{s}} = \pi\,\mathrm{Hz} \approx 3.14\,\mathrm{Hz} \Rightarrow$$ $$x(t) = A \cos(\omega t + \varphi_0) = 0.05 \left(\pi t + \frac{\pi}{4}\right).$$

To find the maximum speed, we need to take the derivative of $x(t)$:

$$V(t) = x'(t)$$ $$V(t) = -A\omega \sin(\omega t + \varphi_0)$$

$-1 \leq \sin(\omega t + \varphi_0) \leq 1 \Rightarrow$ maximum speed when the sine is equal to 1 or $-1$:

$$V_{\max} = -A\omega \cdot (-1) = A\omega = 0.05\,\mathrm{m} \cdot \pi = 0.157\,\frac{\mathrm{m}}{\mathrm{s}}$$

**Answer:** Equation of motion of a simple harmonic oscillator:

$$x(t) = 0.05 \left(\pi t + \frac{\pi}{4}\right). \text{ maximum speed: } V_{\max} = 0.157\,\frac{\mathrm{m}}{\mathrm{s}}.$$