At a mela, walk far away from a merry-go-round and look at it from the side, level with the seats. The circle disappears. Your friend on a wooden horse seems to move along a straight line. The horse slows down at the left end, stops for an instant and comes back. It is fastest in the middle, then slows down again at the right end. This looks very much like the oscillation of Lesson 2.
The shadow of a point on a circle
Let a point P go round a circle of radius \(A\) with a steady angular velocity \(\omega\). Measure its angle \(\theta\) anticlockwise from the horizontal diameter. If the angle was \(\varphi\) at \(t = 0\), then \(\theta = \omega t + \varphi\), as in the chapter Kinematics of Circular Motion. Drop a perpendicular from P on to the vertical diameter, and call its foot Q. Q is the shadow of P on that diameter.
The displacement of Q from the centre O is the height of P above the horizontal diameter. That height is \(A\sin\theta\). So the shadow moves as
\[ x = A\sin(\omega t + \varphi) \]
This is exactly the motion of SHM from Lesson 4. The circle used in this way is called the reference circle.
Why the shadow obeys the law of SHM
The equation for \(x\) already shows SHM, but the arrows in Fig 5.1 show why. The motion of Q is the vertical part of the motion of P. So the velocity and the acceleration of Q are the vertical components of those of P. P moves at the steady speed \(A\omega\) along the tangent, and its acceleration is the centripetal acceleration \(A\omega^2\), directed to O.
\[ x = A\sin\theta \]
Why. the height of P above the horizontal diameter, which is the displacement of its shadow Q.
\[ v_Q = A\omega\cos\theta \]
Why. P's velocity has size \(A\omega\) and lies along the tangent, at the angle \(\theta\) to the vertical. Its vertical component is \(A\omega\cos\theta\), the result of Lesson 4.
\[ a_Q = -A\omega^2\sin\theta \]
Why. P's acceleration has size \(A\omega^2\) and points from P to O. Its vertical component is \(A\omega^2\sin\theta\), pointing down while P is above the horizontal diameter.
\[ a_Q = -\omega^2 x \]
Read it. put \(A\sin\theta = x\) from step 1. The shadow's acceleration is \(-\omega^2\) times its displacement, the law of SHM.
P itself moves at a steady speed. Yet its shadow speeds up towards O and slows down near the ends. The centripetal acceleration always points to O, and its vertical part always points to the mean position of the shadow. The farther the shadow is from O, the larger that vertical part.
The choice of diameter does not matter. The shadow of P on the horizontal diameter is \(A\cos\theta\). That is the same SHM with the phase shifted by \(\pi/2\), as Lesson 4 showed. The friend on the merry-go-round, seen edge-on, is exactly such a shadow. The edge-on motion is SHM, with the radius of the ride as its amplitude and one turn of the ride as its period.
A point moves round a circle at a steady speed, and its shadow on a diameter performs SHM. The speed of the point is doubled, and the radius is kept the same. What happens to the greatest speed of the shadow?
A point goes round the reference circle of an SHM. Which statements are true?
The phasor
Keep only the radius OP of the reference circle and draw it as an arrow. This rotating arrow is called a phasor. Its length is the amplitude \(A\). It turns anticlockwise at the steady rate \(\omega\). Its angle with the horizontal diameter is the phase \(\Phi\). Its projection on the vertical diameter is the displacement \(x\). A single arrow carries all the facts of an SHM.
Phasors are most useful when there are two motions. Take two SHMs with the same \(\omega\) and the same amplitude, but different initial phases. Their phasors turn together, so the angle between them never changes. That angle is the phase difference \(\Delta\varphi\). The motion whose phasor is ahead, anticlockwise, leads. It reaches each peak earlier, by the time \(\Delta\varphi/\omega\).
With \(\Delta\varphi = \pi/2\) in Fig 5.2, SHM 2 reaches its peak a quarter period, \(1\,\text{s}\), before SHM 1. This is the relation between \(\cos\omega t\) and \(\sin\omega t\): the cosine leads by \(\pi/2\). With \(\Delta\varphi = \pi\), the two shadows are always on opposite sides of O. Such motions are said to be in antiphase. With \(\Delta\varphi = 0\) they would move together, in phase.
The same picture explains Lesson 4. The velocity of an SHM, \(A\omega\cos\Phi\), is the projection of an arrow of length \(A\omega\) that is always \(\pi/2\) ahead of the displacement phasor. The acceleration phasor has length \(A\omega^2\) and is \(\pi\) ahead, pointing opposite to the displacement phasor.
Two particles perform the same SHM along one line. At one instant particle 1 is at \(x = +A\) and particle 2 is at \(x = -A\). What is the phase difference between them?
Time as an angle
The reference circle turns a hard question into an easy one. How long does a particle in SHM take to go from one position to another? The shadow moves at a changing speed, so distance divided by speed does not work. The point on the circle, however, sweeps angle at a steady rate \(\omega\). So find the angle it sweeps, and divide by \(\omega\).
A particle in SHM goes from the mean position O to \(x = A/2\). Then it goes on from \(A/2\) to the extreme \(A\). The two distances are equal. How do the two times compare?
The figure above uses an SHM of period \(12\,\text{s}\), so the point sweeps \(30°\) every second. From O to \(A/2\), \(\sin\theta = 1/2\) gives \(\theta = 30°\), a time of \(1\,\text{s}\), or \(T/12\). From \(A/2\) to \(A\), the angle grows from \(30°\) to \(90°\). The \(60°\) take \(2\,\text{s}\), or \(T/6\). From \(-A/2\) to \(+A/2\), the angle runs from \(-30°\) to \(+30°\), again \(60°\) and \(T/6\).
Every position of the shadow has two points above it on the circle, one on the right half and one on the left half. On the right half the shadow moves upwards, towards \(+x\). On the left half it moves downwards. So choose the point on the half that matches the direction of motion, then measure the angle swept between the two points, anticlockwise.
A particle performs SHM with amplitude \(4\,\text{cm}\) and period \(2.4\,\text{s}\). How long does it take to go from \(x = -2\,\text{cm}\) to \(x = +2\sqrt{3} \approx 3.46\,\text{cm}\), moving towards \(+x\) the whole way?
The particle moves towards \(+x\), so both points lie on the right half of the reference circle. The angles satisfy \(\sin\theta = x/A\).
\(\sin\theta_1 = -2/4 = -1/2\), so \(\theta_1 = -30°\). \(\sin\theta_2 = 2\sqrt{3}/4 = \sqrt{3}/2\), so \(\theta_2 = 60°\).
The angle swept is \(60° - (-30°) = 90°\), a quarter of a turn. So \(\Delta t = T/4 = 2.4/4 = 0.6\,\text{s}\).
The particle takes \(0.6\,\text{s}\). The distance is \(5.46\,\text{cm}\), more than the amplitude. Yet the time is only a quarter period, because the particle crosses the mean position at full speed.
For what fraction of each period is a particle in SHM less than \(A/\sqrt{2}\) from its mean position?
\(|x| < A/\sqrt{2}\) means \(|\sin\theta| < 1/\sqrt{2}\). On the reference circle that is within \(45°\) of the horizontal diameter.
There are two such arcs, one on each side, from \(-45°\) to \(+45°\) and from \(135°\) to \(225°\). Each is \(90°\) long.
Together they cover \(180°\), half of the circle. So the particle spends half of each period within \(A/\sqrt{2}\) of the mean position, and half beyond it.
A particle performs SHM with period \(6\,\text{s}\). Starting from the mean position, how long does it take to reach half the amplitude, in seconds?
A particle in SHM of period \(2\,\text{s}\) is at \(x = +A/2\), moving towards the mean position. How long does it take to reach \(x = -A/2\), in seconds?
In January 1610 Galileo turned his new telescope on Jupiter. Beside the planet he saw four small stars, all on one straight line. Night after night they changed places, moving from one side of Jupiter to the other and back again. Galileo understood that they were moons going round Jupiter. We see their orbits almost edge-on, so each moon seems to swing to and fro along a line. A graph of a moon's distance from Jupiter against time is a sine curve. Each moon is the shadow of a circular motion, and the amplitude of its swing is the radius of its orbit.
Some machines must turn a steady rotation into a to-and-fro motion. In a Scotch yoke, a pin fixed on a turning wheel slides inside a straight slot cut in a sliding frame, the yoke. The yoke can move only along one line, so it follows only the component of the pin's motion along that line. The pin is P on the reference circle, and the yoke is its shadow. The yoke therefore moves in exact SHM, with the pin's radius as its amplitude. The mechanism is used in some valve drives and pumps.
- Dividing the distance by the greatest speed to find the time between two positions.The speed of the shadow changes. Use the angle swept on the reference circle: \(\Delta t = \Delta\theta/\omega\).
- Taking the point on the wrong half of the reference circle.The right half means moving towards \(+x\), the left half towards \(-x\). Choose the half that matches the direction of motion.
- Thinking the point on the reference circle speeds up and slows down.The point moves at a steady speed \(A\omega\); only its shadow changes speed.
- Assuming equal distances take equal times in SHM.The particle is slow near the ends. From \(A/2\) to \(A\) takes twice as long as from O to \(A/2\).