Simple Harmonic Motion and Trigonometric Transformations

Problem Statement

Simple Harmonic Motion (SHM) is a cornerstone of mechanics, characterising oscillating systems from pendulums to springs. Mathematically, it is defined by the differential equation $\ddot{x} = -n^2x$, where acceleration is proportional and opposite to displacement.

However, solving SHM problems frequently goes beyond calculus; it requires deep mastery of trigonometric transformations and the ratios of sums and differences of angles. Often, the displacement of a particle is given as a sum of sine and cosine terms, which must be transformed into a single wave function to extract meaningful physical properties like amplitude, phase, and maximum velocity.

Consider a particle moving in Simple Harmonic Motion. Its displacement $x$ (in metres) from the origin $O$ at time $t$ (in seconds) is given by the equation: $x(t) = 3 \sin(2t) + 4 \cos(2t)$

(a) Use the auxiliary angle method to express $x(t)$ in the form $R \cos(2t - \alpha)$, where $R > 0$ and $0 < \alpha < \frac{\pi}{2}$. Give the exact value of $R$ and $\alpha$ to two decimal places.

(b) State the amplitude of the motion.

(c) Find the maximum speed of the particle and determine the first time $t > 0$ when this maximum speed occurs.

Hints

• Part (a): Use the trigonometric compound angle formula $\cos(A - B) = \cos A \cos B + \sin A \sin B$. Expand $R \cos(2t - \alpha)$ and equate the coefficients of $\sin(2t)$ and $\cos(2t)$ with the original equation.
• Part (b): Once the equation is in the form $x(t) = R \cos(2t - \alpha)$, the amplitude is simply the coefficient $R$, as the maximum value of the cosine function is $1$.
• Part (c): Speed is the magnitude of velocity. Find the velocity function $v(t) = \dot{x}(t)$ by differentiating your result from part (a). The maximum speed occurs when the sine function in the velocity equation reaches its peak magnitude.

Solutions

Part (a): Auxiliary Angle Transformation

1. We want to express $3 \sin(2t) + 4 \cos(2t)$ in the form $R \cos(2t - \alpha)$.
2. Expand the target form using the compound angle identity: $R \cos(2t - \alpha) = R (\cos(2t)\cos\alpha + \sin(2t)\sin\alpha)$ $R \cos(2t - \alpha) = (R \cos\alpha)\cos(2t) + (R \sin\alpha)\sin(2t)$
3. Equate coefficients with our original equation $x(t) = 4 \cos(2t) + 3 \sin(2t)$:
   • Coefficient of $\cos(2t)$: $R \cos\alpha = 4$ (Equation 1)
   • Coefficient of $\sin(2t)$: $R \sin\alpha = 3$ (Equation 2)
4. Find $R$ by squaring and adding Equation 1 and Equation 2: $(R \cos\alpha)^2 + (R \sin\alpha)^2 = 4^2 + 3^2$ $R^2 (\cos^2\alpha + \sin^2\alpha) = 16 + 9$ Since $\cos^2\alpha + \sin^2\alpha = 1$, we have: $R^2 = 25 \implies R = 5$ (since $R > 0$).
5. Find $\alpha$ by dividing Equation 2 by Equation 1: $\frac{R \sin\alpha}{R \cos\alpha} = \frac{3}{4}$ $\tan\alpha = \frac{3}{4}$ $\alpha = \arctan(0.75) \approx 0.6435 \text{ radians}$
6. Therefore, the equation of motion is: $x(t) = 5 \cos(2t - 0.64)$

Part (b): Finding Amplitude

1. From the transformed equation $x(t) = 5 \cos(2t - 0.64)$, the amplitude is the maximum displacement from the centre of motion.
2. Since the maximum value of $\cos(\theta)$ is 1, the maximum value of $x(t)$ is $5 \times 1 = 5$.
3. The amplitude is $5$ metres.

Part (c): Maximum Speed and Time

1. Find the velocity by differentiating the displacement function: $v(t) = \frac{d}{dt} [5 \cos(2t - 0.64)]$ $v(t) = 5 \times (-2) \sin(2t - 0.64)$ $v(t) = -10 \sin(2t - 0.64)$
2. Speed is $|v(t)|$. The maximum speed occurs when the magnitude of the sine term is maximum (i.e., when $\sin(2t - 0.64) = \pm 1$).
3. Therefore, Maximum Speed = $|-10 \times (\pm 1)| = 10 \text{ m/s}$.
4. To find the first time $t > 0$ this occurs, we need the sine function to equal $-1$ or $1$. We know that maximum speed in SHM occurs as the particle passes through the origin ($x=0$). Setting $x(t) = 0$: $5 \cos(2t - 0.6435) = 0$ $2t - 0.6435 = \frac{\pi}{2}, \frac{3\pi}{2}, \dots$ For the first time $t > 0$, we take the smallest positive argument: $2t - 0.6435 = \frac{\pi}{2}$ $2t = \frac{\pi}{2} + 0.6435 \approx 1.5708 + 0.6435 = 2.2143$ $t \approx 1.11 \text{ seconds}$

Takeaways

• The Auxiliary Angle Method: This is a vital technique for combining linear combinations of sines and cosines of the same frequency into a single phase-shifted wave. It simplifies finding roots, maxima, and minima.
• Amplitude and Phase: In the form $R \cos(nt - \alpha)$, $R$ immediately gives the amplitude, and $\alpha$ represents the phase shift (the "starting position" of the wave relative to a standard cosine curve).
• SHM Properties: In Simple Harmonic Motion, the maximum speed always occurs at the centre of motion ($x=0$), and the maximum acceleration always occurs at the extremes (amplitude, $x = \pm R$).

Further Readings

• HSC Mechanics: https://vumaths.com/booklets/hsc-mechanics/
• HSC Trigonometry: https://vumaths.com/booklets/hsc-trigonometry/
• HSC Functions: https://vumaths.com/booklets/hsc-functions/

Connect with me

If you want to master the tricks of trigonometric transformations in mechanics, head over to Vu's Maths Hub for my comprehensive HSC booklets. You can also catch me breaking down these exact problems step-by-step on my YouTube channel. Follow my daily maths insights on Instagram and check out my longer-form thoughts on education via my Substack.