use the equations v = v0 cos ωt and v = ± ω ()xx022−

Simple Harmonic Oscillations

Simple harmonic motion (SHM) describes the oscillatory motion of a system where the restoring force is directly proportional to the displacement from equilibrium and acts in the opposite direction. The motion can be described by sinusoidal functions of time.

Key Equations

• Displacement: $x(t) = x_0 \sin(\omega t + \phi)$
• Velocity (using the cosine form): $$v = v_0 \cos(\omega t)$$
• Velocity expressed in terms of displacement: $$v = \pm \,\omega \sqrt{x_0^{2} - x^{2}}$$
• Acceleration: $a(t) = -\omega^{2} x(t)$

Derivation of $v = v_0 \cos(\omega t)$

Starting from the displacement equation $x = x_0 \sin(\omega t + \phi)$, differentiate with respect to time:

$$\frac{dx}{dt}=v = x_0 \omega \cos(\omega t + \phi)$$

If the phase constant $\phi$ is chosen such that at $t = 0$ the particle is at maximum displacement ($x = x_0$), then $\phi = \frac{\pi}{2}$ and the velocity simplifies to:

$$v = v_0 \cos(\omega t)$$

where $v_0 = \omega x_0$ is the maximum speed.

Derivation of $v = \pm \omega \sqrt{x_0^{2} - x^{2}}$

From the energy conservation in SHM, the total mechanical energy $E$ is constant:

$$E = \frac{1}{2}k x^{2} + \frac{1}{2}m v^{2} = \frac{1}{2}k x_0^{2}$$

Using $k = m\omega^{2}$ and solving for $v$ gives:

$$\frac{1}{2}m v^{2} = \frac{1}{2}m\omega^{2}(x_0^{2} - x^{2})$$ $$v^{2} = \omega^{2}(x_0^{2} - x^{2})$$ $$v = \pm \,\omega \sqrt{x_0^{2} - x^{2}}$$

The sign indicates the direction of motion: positive when moving away from the equilibrium point and negative when moving towards it.

Physical Interpretation of the \cdot ariables

Symbol	Quantity	Units	Typical Meaning in SHM
$x$	Displacement from equilibrium	metre (m)	Instantaneous position of the particle
$x_0$	Amplitude	metre (m)	Maximum displacement
$v$	Velocity	metre per second (m s⁻¹)	Rate of change of displacement
$v_0$	Maximum speed	metre per second (m s⁻¹)	$v_0 = \omega x_0$
$a$	Acceleration	metre per second squared (m s⁻²)	$a = -\omega^{2} x$
$\omega$	Angular frequency	radian per second (rad s⁻¹)	$\omega = 2\pi f = \sqrt{k/m}$
$f$	Frequency	hertz (Hz)	$f = \frac{\omega}{2\pi}$
$T$	Period	second (s)	$T = \frac{1}{f} = \frac{2\pi}{\omega}$

Example Problem

Problem: A mass–spring system oscillates with an amplitude of $0.10\ \text{m}$ and an angular frequency of $5\ \text{rad s}^{-1}$. Determine the speed of the mass when it is $0.06\ \text{m}$ from equilibrium.

1. Identify the given quantities: $x_0 = 0.10\ \text{m}$, $\omega = 5\ \text{rad s}^{-1}$, $x = 0.06\ \text{m}$.
2. Use the velocity–displacement relation: $$v = \pm \,\omega \sqrt{x_0^{2} - x^{2}}$$
3. Calculate: $$v = \pm 5 \sqrt{(0.10)^{2} - (0.06)^{2}}$$ $$v = \pm 5 \sqrt{0.0100 - 0.0036}$$ $$v = \pm 5 \sqrt{0.0064}$$ $$v = \pm 5 \times 0.080 = \pm 0.40\ \text{m s}^{-1}$$
4. The magnitude of the speed is $0.40\ \text{m s}^{-1}$. The sign depends on whether the mass is moving towards or away from the equilibrium position.

Common Misconceptions

• Amplitude vs. Maximum speed: Amplitude $x_0$ is a distance, while maximum speed $v_0$ is a velocity. They are related by $v_0 = \omega x_0$, not equal.
• Phase of sine and cosine: The choice of sine or cosine depends on the initial conditions. Both describe SHM; the phase constant $\phi$ adjusts the starting point.
• Sign of the velocity expression: The $\pm$ in $v = \pm \omega \sqrt{x_0^{2} - x^{2}}$ does not imply two different speeds; it indicates direction of motion.

Suggested Diagram

Summary

In simple harmonic oscillations, the velocity can be expressed either as a cosine function of time, $v = v_0 \cos(\omega t)$, or directly in terms of displacement, $v = \pm \omega \sqrt{x_0^{2} - x^{2}}$. Both forms are interchangeable and useful for different problem‑solving strategies. Mastery of these equations, together with a clear understanding of the physical meaning of each symbol, is essential for success in Cambridge A‑Level Physics.