A mason's plumb bob hangs on a long thread at a building site. Pull it aside and let go, and it swings to and fro along a line. Tap it sideways just as it reaches the end of a swing, and it starts to move round an ellipse. Tap it sideways as it passes the middle instead, and it swings along a tilted line.
Two SHMs at right angles
Until now both SHMs acted along the same line. Now let one act along \(x\) and the other along \(y\), at right angles. The body then moves in a plane, and its position vector is \(\vec r = x\,\hat i + y\,\hat j\). Take the same angular frequency \(\omega\) for both: \[ x = A\sin\omega t, \qquad y = B\sin(\omega t + \varphi). \] Here \(A\) and \(B\) are the two amplitudes and \(\varphi\) is the phase of the \(y\) motion ahead of the \(x\) motion. Each coordinate moves on its own, as in any motion in a plane. The question is what path the body traces.
The path is the relation between \(x\) and \(y\) with the time removed. Eliminating \(t\) takes four short steps.
\[ \dfrac{x}{A} = \sin\omega t, \qquad \dfrac{y}{B} = \sin\omega t\cos\varphi + \cos\omega t\sin\varphi \]
Why. divide each equation by its amplitude, and expand the sine of a sum.
\[ \dfrac{y}{B} - \dfrac{x}{A}\cos\varphi = \cos\omega t\,\sin\varphi \]
Why. replace \(\sin\omega t\) by \(x/A\) and move that term to the left.
\[ \left(\dfrac{y}{B} - \dfrac{x}{A}\cos\varphi\right)^2 = \left(1 - \dfrac{x^2}{A^2}\right)\sin^2\varphi \]
Why. square both sides, and write \(\cos^2\omega t = 1 - \sin^2\omega t = 1 - x^2/A^2\).
\[ \dfrac{x^2}{A^2} + \dfrac{y^2}{B^2} - \dfrac{2xy\cos\varphi}{AB} = \sin^2\varphi \]
Read it. expand the left side and collect the \(x^2\) terms, using \(\cos^2\varphi + \sin^2\varphi = 1\). The time has gone.
This is the equation of an ellipse that fits inside the box \(-A \le x \le A\), \(-B \le y \le B\). The phase difference \(\varphi\) decides its shape and its tilt.
A line, an ellipse or a circle
Put \(\varphi = 0\). Then \(\sin\varphi = 0\) and \(\cos\varphi = 1\), and the equation becomes \((x/A - y/B)^2 = 0\). So \(y = (B/A)x\): a straight line through the origin, rising to the right along the diagonal of the box. The two motions are in phase, so the body reaches the corner \((A, B)\) and the corner \((-A, -B)\) in turn. For \(\varphi = \pi\), \(y = -(B/A)x\): a straight line along the other diagonal.
Put \(\varphi = \pi/2\). Then \(\cos\varphi = 0\) and \(\sin\varphi = 1\), and the equation becomes \(x^2/A^2 + y^2/B^2 = 1\). This is an ellipse whose axes lie along \(x\) and \(y\), with semi-axes \(A\) and \(B\). If also \(A = B\), it becomes the circle \(x^2 + y^2 = A^2\).
The circle is uniform circular motion, run backwards from the chapter Kinematics of SHM. There, a point moving round a circle cast an SHM as its shadow. Here two perpendicular SHMs of equal amplitude, a quarter cycle apart, rebuild the circle. The body moves round it at the steady speed \(A\omega\). For every other \(\varphi\) the path is an ellipse tilted in the box.
Watch which way the point goes round. At \(t = 0\), \(x = 0\) and \(y = B\sin\varphi\). For \(\varphi\) between \(0\) and \(\pi\), this is on the upper half of the \(y\) axis, and \(x\) is increasing. So the point moves to the right across the top of the path, which is clockwise. For \(\varphi\) between \(\pi\) and \(2\pi\) the point goes round anticlockwise.
A particle is subject to \(x = A\sin\omega t\) and \(y = A\sin(\omega t + \pi)\) at the same time. What is its path?
A particle moves with \(x = 3\sin\omega t\) and \(y = 4\cos\omega t\), in centimetres. What is its path?
A particle moves with \(x = A\cos\omega t\) and \(y = A\sin\omega t\). Which describes its motion?
Reading the phase difference from an ellipse
The ellipse carries its phase difference in its shape. Where does it cross the \(y\) axis? There \(x = 0\), so \(\sin\omega t = 0\) and \(\cos\omega t = \pm 1\). Then \(y = B\sin(\omega t + \varphi) = \pm B\sin\varphi\). So the ellipse cuts the \(y\) axis at \(\pm B\sin\varphi\), while its highest points are at \(\pm B\).
So \(\sin\varphi = y_0/B\), where \(y_0\) is the height of the cut. A laboratory oscilloscope uses this idea to measure the phase difference between two electrical signals of one frequency. One signal moves the spot across the screen and the other moves it up and down. The ratio of the cut to the half-height of the ellipse gives \(\sin\varphi\). The tilt of the ellipse then says whether \(\varphi\) is below or above \(90°\), as Fig 3.1 shows.
A particle moves with \(x = 2\sin\omega t\) and \(y = 2\sin(\omega t + \pi/3)\), in centimetres. Find the equation of its path and where the path cuts the \(y\) axis.
Here \(A = B = 2\,\text{cm}\) and \(\varphi = 60°\), with \(\cos\varphi = 1/2\) and \(\sin^2\varphi = 3/4\).
The path equation gives \(x^2/4 + y^2/4 - 2xy(1/2)/4 = 3/4\), that is \(x^2 + y^2 - xy = 3\).
At \(x = 0\), \(y^2 = 3\), so the path cuts the \(y\) axis at \(y = \pm\sqrt3 \approx \pm 1.73\,\text{cm}\). As a check, \(B\sin\varphi = 2 \times 0.866 = 1.73\,\text{cm}\).
The equation does not change when \(x\) and \(y\) are swapped, so the ellipse is symmetric about the line \(y = x\). Its long axis lies along that diagonal.
On an oscilloscope, two signals of the same frequency make an ellipse that leans to the right. The ellipse reaches \(3.0\,\text{cm}\) above the centre, and it cuts the vertical axis \(1.5\,\text{cm}\) above the centre. What is the phase difference between the signals, in degrees?
Unequal frequencies: Lissajous figures
Now let the two frequencies differ: \(x = A\sin\omega_x t\) and \(y = B\sin(\omega_y t + \varphi)\). If the ratio \(\omega_x : \omega_y\) is a ratio of small whole numbers, the path closes on itself. After a time equal to a whole number of periods of each motion, the body is back where it started, moving the same way. Such closed paths are Lissajous figures.
Take the ratio \(1 : 2\), so \(x = A\sin\omega t\) and \(y = B\sin(2\omega t + \varphi)\). For \(\varphi = 0\), \(y = 2B\sin\omega t\cos\omega t\). While \(x\) swings once across the box, \(y\) swings up and down twice. The path is a figure of eight lying on its side, crossing itself at the centre. For \(\varphi = \pi/2\), \(y = B\cos 2\omega t = B(1 - 2x^2/A^2)\). The path is then a parabola, which the body runs along and back.
Play the chip \(1 : 2\) in the figure above at \(\varphi = 0°\) and at \(90°\). Then try \(2 : 3\). Count how many times each figure touches the top side of its box, and how many times it touches a side edge. For \(1 : 2\), the figure of eight touches the top twice and each side once. For \(2 : 3\), it touches the top three times and each side twice.
The reason is simple. The figure touches the top once for every oscillation of \(y\), and a side once for every oscillation of \(x\), in one closed round. So the counts give the ratio of the frequencies, as long as the figure does not fold back on itself as the parabola does.
Take a closed Lissajous figure that does not fold back on itself. Count the times it touches a horizontal side of its box, and the times it touches a vertical side. The first count divided by the second equals the frequency of the \(y\) motion divided by that of the \(x\) motion.
A Lissajous figure on an oscilloscope touches the top of its box at two points and each side at one point. What is the ratio of the frequency of the vertical motion to that of the horizontal motion?
An oscilloscope shows two signals at right angles, one of \(50.0\,\text{Hz}\) and the other of \(50.5\,\text{Hz}\). The pattern slowly changes from a line to an ellipse, to a line along the other diagonal, and back, once every \(2\,\text{s}\). Explain why.
The frequencies are nearly equal, so at any instant the pattern looks like the ellipse for equal frequencies with some phase difference. But the faster signal gains on the slower one by \(0.5\) oscillation every second. So the phase difference grows by \(2\pi\) every \(2\,\text{s}\). As it passes \(0\), \(\pi/2\), \(\pi\) and \(3\pi/2\), the pattern goes through a line, an ellipse, the other line and an ellipse again.
The French physicist Lissajous wanted to see sound vibrations. He fixed a small mirror to a prong of each of two tuning forks, set at right angles. A beam of light bounced off one mirror, then the other, and fell on a screen. With both forks sounding, the spot traced a closed curve. From its shape he could compare the two frequencies very precisely and tune one fork to the other. The American Nathaniel Bowditch had drawn the same curves with a pendulum in 1815.
In a physics laboratory an unknown signal goes to the vertical plates of an oscilloscope and a known signal to the horizontal plates. The known frequency is adjusted until the pattern stands still. If the figure touches the top three times and a side once, the unknown frequency is three times the known one. Before digital counters, this was the standard way to check frequencies.
Now the plumb bob. A long thread makes the bob swing with the same period in every direction, so the two swings have equal frequencies. Tapped sideways at the end of a swing, the sideways motion starts at its mean while the first motion is at an extreme. The phase difference is \(\pi/2\), and the path is an ellipse. Tapped at the middle, both motions pass their means together: \(\varphi = 0\), and the path is a tilted line.
- Adding the amplitudes of perpendicular SHMs as if they acted along one line.Motions at right angles do not add to a single SHM. Eliminate \(t\) to find the path in the plane.
- Taking every path of equal frequencies to be a circle.A circle needs equal amplitudes and \(\varphi = \pi/2\). Other cases give a line or an ellipse.
- Reading \(\varphi\) from \(\sin\varphi\) alone.\(\varphi\) and \(180° - \varphi\) give the same cut. The tilt of the ellipse decides between them.