Trigonometric Formulae, Equations and Inequalities

Lessons

Senior 5 · Mathematics

Trigonometric Formulae, Equations and Inequalities

5 lessons

Lesson 5 of 5

Applications: simple harmonic motion and refraction of light

25 min

Formulae at work

The identities of this unit were not invented for exercises. Physicists use them every day for two things you meet in S5 Physics: vibrations, which are described by sines and cosines of time, and light bending as it crosses from air into glass or water, which is described by the sines of angles. This lesson applies the addition formulae, the harmonic form and the methods of solving equations to both.

Definition

Simple harmonic motion (SHM): motion along a line in which the displacement $x$ from a fixed centre at time $t$ is

x = A\sin(\omega t + \varphi), \qquad A \gt 0,\ \omega \gt 0.

Amplitude $A$ : the greatest displacement. Angular frequency $\omega$ (in $\mathrm{rad\,s^{-1}}$ ). Phase constant $\varphi$ : fixes where the motion starts, since $x(0) = A\sin\varphi$ .

Period $T = \frac{2\pi}{\omega}$ (Lesson 1) and frequency $f = \frac{1}{T} = \frac{\omega}{2\pi}$ , in hertz (Hz).

Example: a mass bouncing on a spring, a swinging pendulum (for small swings) and a point on a guitar string all move, to a good approximation, in SHM.

Worked example 1: reading an SHM equation

A mass on a spring moves with $x = 0.05\sin\left(4\pi t + \frac{\pi}{6}\right)$ , with $x$ in metres and $t$ in seconds. Find the amplitude, period and frequency, the displacement at $t = 0$ , and the first time after $t = 0$ at which the mass passes through the centre.

Solution $A = 0.05\,\mathrm{m}$ ; $\omega = 4\pi$ , so $T = \frac{2\pi}{4\pi} = 0.5\,\mathrm{s}$ and $f = 2\,\mathrm{Hz}$ . At $t = 0$ : $x = 0.05\sin\frac{\pi}{6} = 0.025\,\mathrm{m}$ .

The mass is at the centre when $\sin\left(4\pi t + \frac{\pi}{6}\right) = 0$ , that is $4\pi t + \frac{\pi}{6} = k\pi$ . The smallest positive $t$ comes from $k = 1$ : $4\pi t = \frac{5\pi}{6}$ , so $t = \frac{5}{24} \approx 0.208\,\mathrm{s}$ .

Adding two vibrations

When one body takes part in two vibrations of the same angular frequency, such as $x_1 = a\sin\omega t$ and $x_2 = b\cos\omega t$ , the displacements add. Lesson 4 says at once that the result is again simple harmonic, with the same $\omega$ :

x = a\sin\omega t + b\cos\omega t = R\sin(\omega t + \alpha), \qquad R = \sqrt{a^2 + b^2}.

Figure 1. Two vibrations of the same angular frequency add to one vibration of that frequency:

2\sin 5t + 2\sqrt{3}\cos 5t = 4\sin\left(5t + \frac{\pi}{3}\right)

Worked example 2: superposition

A point vibrates with $x_1 = 2\sin 5t$ and $x_2 = 2\sqrt{3}\cos 5t$ at the same time ( $x$ in cm, $t$ in s). Find the resulting motion, its amplitude and period, and the first time it reaches $x = 2\,\mathrm{cm}$ .

Solution $R = \sqrt{4 + 12} = 4$ , $\cos\alpha = \frac{2}{4} = \frac{1}{2}$ and $\sin\alpha = \frac{2\sqrt{3}}{4} = \frac{\sqrt{3}}{2}$ , so $\alpha = \frac{\pi}{3}$ :

x = 4\sin\left(5t + \frac{\pi}{3}\right).

Amplitude $4\,\mathrm{cm}$ ; period $\frac{2\pi}{5} \approx 1.26\,\mathrm{s}$ . For $x = 2$ : $\sin\left(5t + \frac{\pi}{3}\right) = \frac{1}{2}$ , so $5t + \frac{\pi}{3} = \frac{\pi}{6} + 2k\pi$ or $\frac{5\pi}{6} + 2k\pi$ . The first family needs $k \ge 1$ to make $t \gt 0$ ; the second gives $5t = \frac{\pi}{2}$ with $k = 0$ . So the first time is $t = \frac{\pi}{10} \approx 0.314\,\mathrm{s}$ . (The point starts at $x(0) = 2\sqrt{3} \approx 3.46$ , rises to 4 at $t = \frac{\pi}{30}$ , then falls through 2.)

Refraction of light

A ray of light that crosses from one transparent medium into another changes direction. Measure the angles from the normal, the line perpendicular to the surface. If the ray meets the surface at angle $\theta_1$ in a medium of refractive index $n_{1}$ and leaves at angle $\theta_2$ in a medium of index $n_{2}$ , then Snell's law holds:

n_1\sin\theta_1 = n_2\sin\theta_2.

Typical values: air $n \approx 1.00$ , water $n \approx 1.33$ , ordinary glass $n \approx 1.5$ . Going into a denser medium (larger $n$ ), $\sin\theta_2 \lt \sin\theta_1$ : the ray bends towards the normal.

Definition

Angle of incidence $i$ : the angle between the incoming ray and the normal. Angle of refraction $r$ : the angle between the refracted ray and the normal.

Refractive index $n$ of a medium: the ratio $\frac{c}{v}$ of the speed of light in a vacuum to its speed in the medium; it has no unit.

Lateral displacement $d$ : the perpendicular distance between the ray that enters a parallel-sided slab and the ray that leaves it.

Example: from air into glass with $n = 1.50$ at $i = 40^\circ$ : $\sin r = \frac{\sin 40^\circ}{1.50} = 0.4285$ , so $r = 25.37^\circ$ .

The lateral displacement of a slab

At the top face, $\sin i = n\sin r$ . The two faces are parallel, so the ray meets the bottom face at angle $r$ and leaves at angle $i$ : it emerges parallel to the incident ray. In Figure 2, EF is the path in the glass and FG is perpendicular to the extended incident ray. In triangle EGF, the angle at E is $i - r$ and $EF = \frac{t}{\cos r}$ , so

d = EF\sin(i - r) = \frac{t\sin(i - r)}{\cos r}.

Expanding $\sin(i - r)$ with the subtraction formula of Lesson 1 gives a second form, $d = t\,(\sin i - \cos i\tan r)$ .

Figure 2. Drawn to scale for

i = 40^\circ

and

n = 1.50

. The emerging ray is parallel to the incident ray, shifted sideways by

d = F G

. In the right-angled triangle EGF the angle at E is

i - r

and the hypotenuse is

EF = \frac{t}{\cos r}

Worked example 3: a glass block

A ray strikes a glass block of thickness $6.0\,\mathrm{cm}$ and refractive index $1.50$ at an angle of incidence of $40^\circ$ . Find the angle of refraction and the lateral displacement.

Solution $\sin r = \frac{\sin 40^\circ}{1.50} = 0.4285$ , so $r = 25.37^\circ$ and $i - r = 14.63^\circ$ .

d = \frac{6.0 \times \sin 14.63^\circ}{\cos 25.37^\circ} = \frac{6.0 \times 0.2525}{0.9035} = 1.68\,\mathrm{cm} \ \text{(3 s.f.)}.

The ray travels $EF = \frac{6.0}{0.9035} = 6.64\,\mathrm{cm}$ inside the glass.

Important

Critical angle. From glass or water into air, the ray bends away from the normal. It can escape only if $n\sin\theta \le 1$ , that is $\sin\theta \le \frac{1}{n}$ : a trigonometric inequality. The largest such angle $c$ , with $\sin c = \frac{1}{n}$ , is the critical angle. For glass with $n = 1.50$ , $c = \sin^{-1}\frac{1}{1.50} = 41.8^\circ$ ; at larger angles all the light is reflected back inside (total internal reflection), which is how optical fibres carry internet signals.

Warning

Angles are measured from the normal, not from the surface. If a question says the ray makes $30^\circ$ with the surface of the glass, the angle of incidence is $90^\circ - 30^\circ = 60^\circ$ . And set your calculator to degrees when the data are in degrees.

Tip

Check with extremes. If $i = 0$ , then $r = 0$ and $d = 0$ : a ray along the normal goes straight through. If $n = 1$ , then $r = i$ and $d = 0$ : no glass, no shift. A formula that fails these checks is wrong.

Khan Academy Physics: Phase constant (about 10 min). Watch how the constant inside the sine slides the graph of an oscillator left or right, and how to read it from the starting position.

✦

Practice

Give answers to 3 significant figures where they are not exact. The last two are harder.

A particle moves with $x = 3\sin\left(2t - \frac{\pi}{4}\right)$ cm. State the amplitude, the period and the frequency, and find $x$ at $t = 0$ .
For the mass of Worked example 1, find every time $t$ in $[0, 0.5)$ s at which $x = 0.025\,\mathrm{m}$ .
Write $x = 6\sin 2t + 8\cos 2t$ in the form $R\sin(2t + \alpha)$ and find the first time $t \gt 0$ at which $x$ is greatest.
Light passes from air into water ( $n = 1.33$ ) with an angle of incidence of $60^\circ$ . Find the angle of refraction.
A slab $4.0\,\mathrm{cm}$ thick has $n = 1.60$ . Find the lateral displacement of a ray with $i = 50^\circ$ .
Show that $\frac{t\sin(i - r)}{\cos r} = t\,(\sin i - \cos i\tan r)$ .
Find the critical angle for light going from water ( $n = 1.33$ ) into air, and state the angles of incidence for which light escapes from the water.
A point takes part in $x_1 = 5\sin\pi t$ and $x_2 = 5\sin\left(\pi t + \frac{2\pi}{3}\right)$ . Use a sum-to-product formula to show that the resulting motion has amplitude 5.

Show answersHide answers

$A = 3\,\mathrm{cm}$ , $T = \frac{2\pi}{2} = \pi \approx 3.14\,\mathrm{s}$ , $f = \frac{1}{\pi} \approx 0.318\,\mathrm{Hz}$ ; $x(0) = 3\sin\left(-\frac{\pi}{4}\right) = -\frac{3\sqrt{2}}{2} \approx -2.12\,\mathrm{cm}$ .
$\sin\left(4\pi t + \frac{\pi}{6}\right) = \frac{1}{2}$ : $4\pi t + \frac{\pi}{6} = \frac{\pi}{6} + 2k\pi$ or $\frac{5\pi}{6} + 2k\pi$ , so $t = \frac{k}{2}$ or $t = \frac{1}{6} + \frac{k}{2}$ . In $[0, 0.5)$ : $t = 0$ and $t = \frac{1}{6} \approx 0.167\,\mathrm{s}$ .
$R = 10$ , $\tan\alpha = \frac{8}{6}$ , $\alpha = 0.927$ : $x = 10\sin(2t + 0.927)$ . Greatest when $2t + 0.927 = \frac{\pi}{2}$ , so $t = 0.322\,\mathrm{s}$ .
$\sin r = \frac{\sin 60^\circ}{1.33} = 0.651$ , $r = 40.6^\circ$ .
$\sin r = \frac{\sin 50^\circ}{1.60}$ , $r = 28.6^\circ$ ; $d = \frac{4.0\sin 21.4^\circ}{\cos 28.6^\circ} = 1.66\,\mathrm{cm}$ .
$\frac{t(\sin i\cos r - \cos i\sin r)}{\cos r} = t\left(\sin i - \cos i\,\frac{\sin r}{\cos r}\right) = t(\sin i - \cos i\tan r)$ .
$\sin c = \frac{1}{1.33}$ , $c = 48.8^\circ$ . Light escapes for angles of incidence $0^\circ \le \theta \lt 48.8^\circ$ .
$x = 5\left[\sin\pi t + \sin\left(\pi t + \frac{2\pi}{3}\right)\right] = 10\sin\left(\pi t + \frac{\pi}{3}\right)\cos\frac{\pi}{3} = 5\sin\left(\pi t + \frac{\pi}{3}\right)$ , amplitude 5.

1Formulae at work

2Adding two vibrations

3Refraction of light

4The lateral displacement of a slab

Formulae at work

Adding two vibrations

Refraction of light

The lateral displacement of a slab