One equation with named parts
The next lesson proves that every simple harmonic motion can be written in one form:
\[ x = A\sin(\omega t + \varphi) \]
You can already check that this form obeys the law of SHM. Differentiate it twice, as in Lesson 1. The result is \(-\omega^2 A\sin(\omega t + \varphi)\), which is \(-\omega^2 x\). This lesson gives each letter of the equation its name and its meaning.
Amplitude
The amplitude \(A\) is the greatest displacement of the particle from the mean position. Since a sine never exceeds \(1\), \(x\) never exceeds \(A\). The particle moves between \(x = +A\) and \(x = -A\). The amplitude is always taken as positive, and it is measured in metres. It depends on how the motion was started: a harder push gives a larger amplitude.
Period and frequency
The period \(T\) is the time taken for one complete oscillation. Start the clock with the particle at some point, moving in some direction. The period is the time until it is again at that point, moving in the same direction. Passing the same point in the opposite direction does not count, because that state is different.
One period splits into four equal quarters in SHM. From the mean position to \(+A\) takes \(T/4\). Back to the mean position takes another \(T/4\), and the trip to \(-A\) and back takes the last two quarters. So a particle takes \(T/2\) to go from one extreme to the other.
The frequency \(f\) is the number of complete oscillations made in one second. If one oscillation takes \(T\) seconds, then \(1/T\) oscillations take one second. The unit of frequency is s⁻¹, called the hertz, Hz. A tuning fork marked \(256\,\text{Hz}\) makes \(256\) oscillations every second, and each one takes \(1/256\) of a second.
A particle performs SHM with period \(T\). How long does it take to go from the mean position to an extreme position?
The phase
The angle inside the sine, \(\Phi = \omega t + \varphi\), is called the phase of the motion. Its unit is the radian. The phase fixes the displacement \(x = A\sin\Phi\) at each instant. It also fixes the direction of motion. Two instants with the same displacement can have opposite velocities, and the phase tells them apart.
Fig 3.1 lists the four landmarks. At \(\Phi = 0\) the particle is at the mean position, moving towards \(+A\). At \(\Phi = \pi/2\) it is at \(+A\), at rest for an instant. At \(\Phi = \pi\) it is at the mean position again, now moving towards \(-A\). At \(\Phi = 3\pi/2\) it is at \(-A\), at rest. At \(\Phi = 2\pi\) every quantity is as it was at \(\Phi = 0\).
Consider \(\Phi = \pi/6\) and \(\Phi = 5\pi/6\). Both give \(x = A\sin\Phi = A/2\). At \(\pi/6\) the phase lies before \(\pi/2\), so the particle is still on its way out to \(+A\). At \(5\pi/6\) the phase lies after \(\pi/2\), so the particle is on its way back. One number, the phase, holds both facts.
What is the speed of a particle in SHM when its phase is \(3\pi/2\)?
Angular frequency and period
As time passes, the phase grows steadily. Its rate of growth is the constant \(\omega\), since \(d\Phi/dt = \omega\). This is why \(\omega\) is called the angular frequency: it is the number of radians of phase gained each second. The motion repeats when the phase has grown by \(2\pi\). That takes one period, so \(\omega T = 2\pi\).
Notice what the period does not contain. \(T = 2\pi/\omega\) has no \(A\) in it. The constant \(\omega\) comes from the law \(a = -\omega^2 x\), not from the size of the motion. So a large oscillation and a small one of the same system take exactly the same time. A motion whose period does not depend on its amplitude is called isochronous.
A particle in SHM makes \(20\) complete oscillations in \(8\,\text{s}\). What is its angular frequency, in rad/s?
Your heart beats about \(72\) times a minute. Treat the beat as a periodic motion and estimate its angular frequency, in rad/s.
The initial phase
At \(t = 0\) the phase is \(\Phi = \varphi\). So the constant \(\varphi\) is the initial phase, also called the phase constant. It says where the particle was, and which way it moved, when the clock was started. Two particles with the same \(A\) and \(\omega\) but different \(\varphi\) perform the same motion, started at different points of the cycle.
In Fig 3.2, \(\varphi = 0\) starts the particle at the mean position, moving up. \(\varphi = \pi/2\) starts it at \(+A\), at rest. Then \(x = A\sin(\omega t + \pi/2) = A\cos\omega t\) by the rule of Lesson 1. \(\varphi = \pi\) starts it at the mean position moving down, and \(\varphi = 3\pi/2\) starts it at \(-A\).
To find \(\varphi\) from a given start, use two facts. The displacement gives \(\sin\varphi = x_0/A\), where \(x_0\) is the displacement at \(t = 0\). Two angles between \(0\) and \(2\pi\) have this sine. The direction of motion at \(t = 0\) picks one of them: a phase before \(\pi/2\) or after \(3\pi/2\) means the particle moves towards \(+x\).
A particle moves as \(x = 0.03\sin(4\pi t + \pi/6)\), with \(x\) in metres and \(t\) in seconds. Find the amplitude, the angular frequency, the frequency, the period and the initial phase. Where is the particle at \(t = 0\), and which way is it moving?
Compare with \(x = A\sin(\omega t + \varphi)\). The amplitude is \(A = 0.03\,\text{m} = 3\,\text{cm}\), and \(\omega = 4\pi\,\text{rad/s}\).
\(f = \omega/2\pi = 2\,\text{Hz}\), so \(T = 1/f = 0.5\,\text{s}\).
The initial phase is \(\varphi = \pi/6\). At \(t = 0\), \(x = 0.03\sin(\pi/6) = 0.015\,\text{m} = 1.5\,\text{cm}\).
The phase \(\pi/6\) lies before \(\pi/2\), so the particle is on its way out to \(+A\). It moves towards \(+x\).
A particle in SHM is written as \(x = A\sin(\omega t + \varphi)\). At \(t = 0\) it is at \(x = -A\). What is its initial phase \(\varphi\), between \(0\) and \(2\pi\)?
A particle moves as \(x = A\sin(\omega t + \varphi)\). Which statements are true?
In 1583, the story goes, a young medical student named Galileo sat in the cathedral at Pisa. A lamp hanging from the high roof had been set swinging. Galileo timed its swings against his own pulse. As the swings died away and grew smaller, each one still seemed to take the same time. The story was written down by his pupil Vincenzo Viviani many years later. Galileo went on to test the idea with pendulums of his own. For small swings he was right: the period does not depend on the amplitude, exactly as \(T = 2\pi/\omega\) says.
Inside a quartz watch is a tiny tuning fork cut from a quartz crystal. An electric circuit keeps it oscillating at \(32\,768\,\text{Hz}\), so its angular frequency is \(\omega = 2\pi \times 32\,768 \approx 2.06 \times 10^5\,\text{rad/s}\), and one period lasts about \(30.5\) microseconds. The number \(32\,768\) is \(2^{15}\). The circuit halves the frequency fifteen times, and what is left is one pulse each second. That pulse moves the second hand.
- Counting a pass through the same point in the opposite direction as a full period.One period brings back both the position and the direction of motion.
- Confusing \(f\) with \(\omega\).\(f\) counts oscillations per second; \(\omega = 2\pi f\) counts radians of phase per second.
- Thinking that a larger amplitude takes a longer time.In SHM, \(T = 2\pi/\omega\) does not contain \(A\). A bigger swing is also a faster one.
- Finding \(\varphi\) from the displacement alone.\(\sin\varphi = x_0/A\) has two answers between \(0\) and \(2\pi\). Use the direction of motion at \(t = 0\) to choose.