Sit on the sea wall at Marine Drive in Mumbai for a day. In the morning the water laps high against the stones. By early afternoon it has fallen far below them. By evening it is high again. The level rises and falls smoothly, and the pattern repeats about every 12 hours and 25 minutes. A fisherman who knows this pattern can say when the water will be high tomorrow.
Radians
Every equation of this unit measures angles in radians. The chapter Kinematics of Circular Motion defined the radian as the angle whose arc equals the radius. A full turn is \(2\pi\) rad, so half a turn gives \(\pi\) rad \(= 180°\). To turn an angle of \(x\) degrees into \(\theta\) radians, multiply by \(\pi/180\).
\[ \theta = \frac{\pi\,x}{180} \]
So \(60° = \pi/3 \approx 1.047\,\text{rad}\), and \(270° = 3\pi/2 \approx 4.712\,\text{rad}\). One radian is about \(57.3°\). A calculator must be set to radians before it evaluates \(\sin 2\) or \(\cos 0.5\). In degree mode it silently gives a wrong answer.
The graphs of sine and cosine
Plot \(y = \sin\theta\) with \(\theta\) in radians. The curve starts at \(0\) and rises to \(+1\) at \(\theta = \pi/2\). It returns to \(0\) at \(\pi\), falls to \(-1\) at \(3\pi/2\) and is back at \(0\) at \(2\pi\). After \(2\pi\) the same shape repeats for ever. The graph of \(y = \cos\theta\) has exactly the same shape. It starts at its top, \(+1\), at \(\theta = 0\).
Both curves stay between \(-1\) and \(+1\). Both repeat every \(2\pi\), so \(2\pi\) is their period in the angle. A curve of this shape is called a sinusoid, whether it is a sine or a cosine. Every oscillation in this unit is drawn with it.
The graph of \(\sin\theta\) is not a picture of anything moving along a wavy path. It is a graph of one number against another. When a particle's position follows a sine curve in time, the particle itself moves along a straight line, to and fro.
Adding a constant to the angle
Now draw \(y = \sin(\theta + \varphi)\), where \(\varphi\) is a fixed angle. The curve has the value that \(\sin\theta\) had a little later, at \(\theta + \varphi\). So every feature of the curve arrives earlier by \(\varphi\). The whole graph slides to the left by \(\varphi\) without changing its shape. The constant \(\varphi\) is called a phase added to the angle.
Set \(\varphi = \pi/2\). The peak that was at \(\pi/2\) moves to \(\theta = 0\), and the solid curve lies exactly on \(\cos\theta\). Set \(\varphi = \pi\). The curve is now upside down: every value of \(\sin\theta\) has changed its sign. At \(\varphi = 2\pi\) the curve has slid by one full period and looks as it did at the start.
One more rule is used often. The complement of an angle turns a sine into a cosine: \(\sin(\pi/2 - \theta) = \cos\theta\). You may check it on a right-angled triangle. The side opposite one acute angle is the side adjacent to the other.
Which expression equals \(\sin(\theta + \pi)\) for every angle \(\theta\)?
Which of these expressions equal \(\cos\theta\) for every angle \(\theta\)?
A sine plus a cosine
Motion in SHM is often built from two parts, one written with a sine and one with a cosine. Their sum \(a\sin\theta + b\cos\theta\) looks like a new kind of curve. It is not. The sum is always one sinusoid of the same period, shifted by some phase. Try it in the figure below before reading the proof.
\[ R\sin(\theta + \delta) = R\cos\delta\,\sin\theta + R\sin\delta\,\cos\theta \]
Why. expand with the rule \(\sin(A + B) = \sin A\cos B + \cos A\sin B\).
\[ R\cos\delta = a, \qquad R\sin\delta = b \]
Why. the two sides agree for every \(\theta\) only if the coefficients of \(\sin\theta\) and of \(\cos\theta\) agree separately.
\[ R^2(\cos^2\delta + \sin^2\delta) = a^2 + b^2 \;\Rightarrow\; R = \sqrt{a^2 + b^2} \]
Why. square the two equations of step 2 and add them. The bracket is \(1\) for every angle.
\[ \tan\delta = \dfrac{b}{a} \]
Read it. divide the second equation of step 2 by the first. This fixes the phase \(\delta\).
A sine never exceeds \(1\), so \(R\sin(\theta + \delta)\) never exceeds \(R\). The greatest value of \(a\sin\theta + b\cos\theta\) is therefore \(\sqrt{a^2 + b^2}\). It is not \(a + b\), because the two parts never reach their peaks at the same angle. The sine peaks at \(\pi/2\) and the cosine at \(0\).
Write \(3\sin\theta + 4\cos\theta\) as a single sine. State its largest value and the angle at which it first occurs.
Here \(a = 3\) and \(b = 4\), so \(R = \sqrt{9 + 16} = 5\) and \(\tan\delta = 4/3\), giving \(\delta \approx 53.1° \approx 0.927\,\text{rad}\).
So \(3\sin\theta + 4\cos\theta = 5\sin(\theta + 0.927)\).
The largest value is \(5\). It occurs where \(\theta + \delta = \pi/2\), that is at \(\theta = \pi/2 - 0.927 \approx 0.644\,\text{rad}\), or about \(36.9°\).
Adding the two separate peaks, \(3 + 4 = 7\), would be wrong: at \(\theta = 36.9°\) the parts are \(3 \times 0.6 = 1.8\) and \(4 \times 0.8 = 3.2\).
The displacement of a particle is \(x = 5\sin\theta + 12\cos\theta\) centimetres, where the angle \(\theta\) grows steadily. What is the largest displacement, in cm?
Derivatives of sine and cosine
In this unit the position of a particle will be a sine of time. Its velocity \(v = dx/dt\) and its acceleration \(a = dv/dt\) are then derivatives of a sine. Look at the graph of \(\sin t\). Its slope is largest, \(+1\), where the curve crosses zero going up. The slope is zero at the top. This pattern of slopes is exactly the graph of \(\cos t\).
\[ \frac{d}{dt}\sin t = \cos t, \qquad \frac{d}{dt}\cos t = -\sin t \]
The minus sign in the second rule also comes from the slopes. Just after \(t = 0\), \(\cos t\) falls from its top, so its slope is negative while \(\sin t\) is positive.
The chain rule
The time in SHM always appears multiplied by a constant \(\omega\), as in \(\sin\omega t\). To differentiate such a function, use the chain rule. Let \(u = \omega t\). The sine changes with \(u\) at the rate \(\cos u\), and \(u\) changes with \(t\) at the rate \(\omega\). The two rates multiply.
Differentiate \(\sin\omega t\) twice. The first derivative is \(\omega\cos\omega t\). The second is \(-\omega^2\sin\omega t\), which is \(-\omega^2\) times the function we started with. The same happens with \(\cos\omega t\). This property will define simple harmonic motion in the next lesson. Few functions have it, and sine and cosine are the ones that matter.
A quantity changes with time as \(x = 0.2\sin 5t\). Find \(dx/dt\) and \(d^2x/dt^2\), and show that \(d^2x/dt^2 = -25x\).
By the chain rule with \(\omega = 5\), \(dx/dt = 0.2 \times 5\cos 5t = 1.0\cos 5t\).
Differentiating again, \(d^2x/dt^2 = 1.0 \times 5 \times (-\sin 5t) = -5\sin 5t\).
Since \(x = 0.2\sin 5t\), we have \(-5\sin 5t = -25 \times 0.2\sin 5t = -25x\). The factor \(25\) is \(\omega^2\).
What is the derivative of \(\cos 3t\) with respect to \(t\)?
A quantity varies as \(y = 4\sin 2t\). What is \(d^2y/dt^2\)?
Back to the tide
The water at Marine Drive rises and falls about a mean level. Its height \(h\) above that level follows a sinusoid in time quite closely: \(h = a\sin(2\pi t/T + \varphi)\), with \(T\) about \(12.4\) hours. The constant \(a\) is the greatest rise. The phase \(\varphi\) records where in its cycle the water was when the clock read zero. Tomorrow's tide table is today's curve, slid along by the phase.
In 499 CE the astronomer Aryabhata of Kusumapura, near today's Patna, finished the Aryabhatiya. In it he gave a table of the half-chords of a circle, which he called jya, for angles in steps of \(3.75°\). The half-chord of a circle is exactly our sine, times the radius. Arab scholars translated the word as jiba. Later it was misread as jaib, meaning a fold or a pocket. A Latin translator of the twelfth century turned that into sinus. The function that describes every oscillation still carries this Indian name.
The voltage at a wall socket in India is a sinusoid in time, \(V = 325\sin(100\pi t)\) volts. It repeats \(50\) times a second, so its angle grows by \(2\pi \times 50 = 100\pi \approx 314\) radians every second. The peak of \(325\,\text{V}\) is \(\sqrt{2}\) times the \(230\,\text{V}\) printed on appliances. The rate at which this voltage changes is \(dV/dt = 325 \times 100\pi\cos(100\pi t)\), by the chain rule. That rate is what a transformer responds to.
- Evaluating \(\sin\omega t\) with the calculator in degree mode.In every SHM formula the angle is in radians. Switch the calculator to radians first.
- Writing the largest value of \(a\sin\theta + b\cos\theta\) as \(a + b\).The parts never peak together. Use \(R = \sqrt{a^2 + b^2}\).
- Forgetting the factor \(\omega\) when differentiating \(\sin\omega t\).By the chain rule, every differentiation of \(\sin\omega t\) or \(\cos\omega t\) brings one factor \(\omega\).
- Losing the minus sign in the derivative of cosine.\(d(\cos t)/dt = -\sin t\): a cosine falls just after \(t = 0\).