Two children sit on two identical swings in a park in Pune. One swing is pulled back a little, the other a little more, and both are let go together. The two swings reach the ends of their arcs at the same moments, again and again. Next time, one child is let go a moment after the other. Now one swing is always at an end of its arc while the other passes the lowest point.
Three numbers fix an SHM
In the chapter Kinematics of SHM, one simple harmonic motion was written as \(x = A\sin(\omega t + \varphi)\). Here \(x\) is the displacement from the mean position and \(A\) is the amplitude. The angular frequency \(\omega\) sets the period, \(T = 2\pi/\omega\). The initial phase \(\varphi\) says where in its cycle the particle is at \(t = 0\). These three numbers fix the motion completely.
Now take two particles in SHM along the same line, about the same mean position. They can differ in \(A\), in \(\omega\), in \(\varphi\), or in all three. To see what each number does, this lesson changes one number at a time and keeps the other two equal.
Same period, different amplitudes
Let the two motions be \(x_1 = A_1\sin\omega t\) and \(x_2 = A_2\sin\omega t\), with the same \(\omega\) and the same phase. Divide one by the other: \(x_2/x_1 = A_2/A_1\) at every instant. So the two displacements are always in the same ratio. Both are zero at the same instants, and both reach their extremes at the same instants.
The velocities follow by differentiating: \(v = A\omega\cos\omega t\) for each particle. So \(v_2/v_1 = A_2/A_1\) as well. The particle with the larger amplitude covers a longer path in the same time, so it must move faster. Its greatest speed \(A\omega\) and its greatest acceleration \(A\omega^2\) are both larger, in the ratio of the amplitudes.
Play the first chip of Fig 1.1. Particle 2 swings twice as far as particle 1, but the two dots cross the dashed line together and turn back together. The two graphs cross the \(t\) axis at the same points and peak at the same instants. Only the heights differ.
Two particles oscillate in phase with the same period. Their amplitudes are \(2\,\text{cm}\) and \(5\,\text{cm}\). At the instant the first particle is at \(x = 1.2\,\text{cm}\), where is the second particle, in cm?
Same amplitude, different periods
Now let \(x_1 = A\sin\omega_1 t\) and \(x_2 = A\sin\omega_2 t\), with \(\omega_2\) larger than \(\omega_1\). The period \(T_2 = 2\pi/\omega_2\) is then shorter than \(T_1\). In every period a particle covers a path of \(4A\): out to one extreme, back through the mean, out to the other extreme and back. Particle 2 covers this same \(4A\) in less time, so it moves faster on average.
Its greatest speed is \(A\omega_2\), larger than \(A\omega_1\). Its greatest acceleration \(A\omega_2^2\) is larger still, because \(\omega\) is squared. A short period therefore means a violent motion even for a small amplitude. The greatest speed depends on \(A\) and \(T\) together, as \(v_{max} = 2\pi A/T\).
Particle P oscillates with amplitude \(2\,\text{cm}\) and period \(0.5\,\text{s}\). Particle Q oscillates with amplitude \(4\,\text{cm}\) and period \(2\,\text{s}\). Which has the larger greatest speed, and which the larger greatest acceleration?
For P, \(\omega = 2\pi/0.5 = 4\pi\,\text{rad/s}\). Then \(v_{max} = A\omega = 2 \times 4\pi \approx 25.1\,\text{cm/s}\) and \(a_{max} = A\omega^2 = 2 \times 16\pi^2 \approx 316\,\text{cm/s}^2\).
For Q, \(\omega = 2\pi/2 = \pi\,\text{rad/s}\). Then \(v_{max} = 4\pi \approx 12.6\,\text{cm/s}\) and \(a_{max} = 4\pi^2 \approx 39.5\,\text{cm/s}^2\).
P has half the amplitude of Q, yet twice the greatest speed and eight times the greatest acceleration. Its period is four times shorter, and the period enters the acceleration squared.
Play the second chip of Fig 1.1. Both particles start at the mean position, moving up together. Particle 2 has a period of \(1\,\text{s}\) and particle 1 a period of \(2\,\text{s}\). After \(0.5\,\text{s}\), particle 2 is back at the mean, moving down, while particle 1 is at its top. The two are out of step. At \(t = 2\,\text{s}\), particle 1 has made one oscillation and particle 2 exactly two. Both are again at the mean, moving up, and the pattern repeats.
So two SHMs with different periods are together again only after a time that is a whole number of periods of each. The shortest such time is the least common multiple of the two periods. Between those instants, how far one leads the other changes all the time. The difference of their phases is \((\omega_2 - \omega_1)t\), and it grows steadily with \(t\).
Two particles start together at the mean position, both moving in the positive direction. Their periods are \(0.4\,\text{s}\) and \(0.6\,\text{s}\). After how many seconds are they next together at the mean position, moving in the positive direction?
Two particles have the same amplitude. Particle 1 has period \(2\,\text{s}\) and particle 2 has period \(1\,\text{s}\). Which statements are true?
Same period, different phases
Last, let the amplitude and the period be equal, and let only the initial phase differ: \(x_1 = A\sin(\omega t + \varphi_1)\) and \(x_2 = A\sin(\omega t + \varphi_2)\). The phase of each motion at time \(t\) is the whole angle \(\omega t + \varphi\). Subtract one phase from the other. The \(\omega t\) cancels, and what is left is \(\varphi_2 - \varphi_1\), the same at every instant.
The phase difference of two SHMs of the same angular frequency is \(\Delta\varphi = (\omega t + \varphi_2) - (\omega t + \varphi_1) = \varphi_2 - \varphi_1\). It is constant in time. The motion with the larger phase is said to lead the other.
Three values of \(\Delta\varphi\) have names. At \(\Delta\varphi = 0\) the two motions are in phase: the particles move together, and each reaches its extremes at the same instants as the other. At \(\Delta\varphi = \pi\) they are in opposite phase: one is at \(+A\) exactly when the other is at \(-A\), and they always move in opposite directions. At \(\Delta\varphi = \pi/2\) one leads by a quarter cycle. When one is at an extreme, the other is passing through the mean.
The phases repeat every \(2\pi\), so \(\Delta\varphi = 2\pi\) is the same as \(\Delta\varphi = 0\). A lead of \(3\pi/2\) is the same as a lag of \(\pi/2\). The phase difference is therefore usually quoted between \(0\) and \(\pi\), with a word to say which motion leads.
A lead in angle is a lead in time
The leading particle reaches every stage of its cycle earlier. How much earlier? Particle 2 has phase \(\omega t + \varphi_2\). Particle 1 reaches the same phase a time \(\Delta t\) later, when \(\omega(t + \Delta t) + \varphi_1 = \omega t + \varphi_2\). So \(\omega\,\Delta t = \Delta\varphi\).
A whole period corresponds to a whole turn, \(2\pi\). So a phase difference of \(\pi/2\) is a quarter of a period, and \(\pi\) is half a period. In the park, one swing reaching an end while the other passes the lowest point is a lead of a quarter period.
Two particles execute SHM with the same amplitude and the same frequency. One particle is at its positive extreme at the instant the other is at the mean position. What is the phase difference between them?
Seeing the phase difference as an angle
The chapter Kinematics of SHM drew an SHM as the shadow of a rotating arrow, the phasor. A phasor of length \(A\) turns anticlockwise at the angular speed \(\omega\). At time \(t\) it makes the angle \(\omega t + \varphi\) with the \(x\) axis, so the height of its tip is \(A\sin(\omega t + \varphi)\). That height is the displacement of the SHM.
Draw both phasors on one diagram, as in Fig 1.2. They turn at the same rate, so the angle between them never changes. That angle is the phase difference. At \(\Delta\varphi = 0\) the two arrows lie on top of each other. At \(\Delta\varphi = \pi\) they point in opposite directions, so the two tips are always at opposite heights. At \(\pi/2\) they are at right angles. When one arrow lies flat, the other points straight up or down. So one particle is at the mean while the other is at an extreme.
Two particles oscillate with amplitude \(4\,\text{cm}\) and period \(2\,\text{s}\) about the same mean position. At one instant both are at \(x = +2\,\text{cm}\). Particle 1 is moving away from the mean, and particle 2 is moving towards it. Find the phase difference and the time by which one leads the other.
Write \(x = A\sin\Phi\) and \(v = A\omega\cos\Phi\), where \(\Phi\) is the phase at this instant. The position fixes \(\sin\Phi\), and the direction of motion fixes the sign of \(\cos\Phi\).
For both, \(\sin\Phi = 2/4 = 1/2\). Particle 1 has \(v > 0\), so \(\cos\Phi_1 > 0\). Particle 2 has \(v < 0\), so \(\cos\Phi_2 < 0\).
\(\Phi_1 = \pi/6\) and \(\Phi_2 = 5\pi/6\), so \(\Delta\varphi = 5\pi/6 - \pi/6 = 2\pi/3\). With \(\omega = 2\pi/2 = \pi\,\text{rad/s}\), \(\Delta t = \Delta\varphi/\omega = 2/3\,\text{s} \approx 0.67\,\text{s}\).
Particle 2 leads particle 1 by \(2\pi/3\), that is \(120°\). It reaches every stage of its cycle about \(0.67\,\text{s}\) earlier. The last chip of Fig 1.2 shows this pair.
Two SHMs have the same period, \(2\,\text{s}\). The second particle reaches each extreme \(0.25\,\text{s}\) before the first. What is the phase difference between them, in degrees?
Two particles oscillate in opposite phase with the same amplitude \(A\). At one instant particle 1 is at \(x = +A/2\), moving towards \(+A\). Where is particle 2, and which way is it moving?
A phase difference is a fixed number only for two SHMs of the same frequency. For different frequencies the difference of the phases, \((\omega_2 - \omega_1)t + \varphi_2 - \varphi_1\), changes with time. Saying "they differ in phase by \(\pi/3\)" then has meaning only at one instant.
Huygens had invented the pendulum clock. In February 1665, ill in bed, he watched two of his clocks that hung from the same wooden beam. Their pendulums always swung in opposite directions: they kept a phase difference of \(\pi\). He disturbed one, and within about half an hour they had returned to opposite phase. He called it "an odd kind of sympathy". The tiny shaking of the shared beam coupled the two pendulums.
Many two-cylinder motorcycle engines are built so that their two pistons move in opposite phase. One piston rises while the other falls, each in a motion close to simple harmonic. The forces that the moving pistons exert on the frame then nearly cancel. The engine shakes the rider much less than a single large piston would.
Return to the swings in the park. Released together, the two swings are in phase: the larger swing moves faster, but both reach the ends of their arcs together. Released a moment apart, they differ in phase. When one swing is at an end while the other passes the lowest point, the phase difference is \(\pi/2\). The later swing lags by a quarter of its period.
- Quoting a fixed phase difference for two SHMs of different periods.The difference of the phases changes with time unless the periods are equal.
- Deciding the phase from the position alone.Two phases give the same \(x\). Use the direction of motion, the sign of \(\cos\Phi\), to choose between them.
- Thinking that the larger amplitude always means the larger greatest speed.\(v_{max} = 2\pi A/T\) depends on the period too. A small, fast swing can outrun a large, slow one.