A block on a spring oscillates between \(x = -4\,\text{cm}\) and \(x = +4\,\text{cm}\), once every \(2.4\,\text{s}\). It moves from \(x = -2\,\text{cm}\) to \(x = +3.46\,\text{cm}\) without turning back. That stretch is \(5.46\,\text{cm}\), longer than the amplitude. How long does the block take? The speed of the block changes at every instant, so the distance divided by one speed cannot give the time. Something steadier is hidden inside the motion.
An arm that turns
Take an arm OP of length \(A\), pinned at the point O. Let the arm turn anticlockwise at a steady rate. The angle it turns in each second is called the angular frequency \(\omega\), measured in radians per second. Measure the angle \(\theta\) of the arm from the horizontal line through O. If the angle was \(\phi\) at \(t = 0\), then after a time \(t\) the angle is \(\theta = \omega t + \phi\).
Draw a perpendicular from the tip P to the vertical line through O. The foot of this perpendicular is the shadow of P on the vertical line. The height of the shadow above O is one side of a right triangle whose hypotenuse is the arm. By the definition of the sine, this height is \(A\sin\theta\). So the shadow moves as
\[ y = A\sin(\omega t + \phi) \]
Watch Fig 1.1 closely. The arm turns at a perfectly steady rate, but its height does not change steadily. The height changes fastest when the arm crosses the horizontal line. It hardly changes when the arm points straight up or straight down. A steady turning produces an uneven rise and fall, and that rise and fall is the shape of the sine curve.
A sinusoid is a quantity that varies with time as \(y = A\sin(\omega t + \phi)\), where \(A\), \(\omega\) and \(\phi\) are constants. The phasor of the sinusoid is an arrow of length \(A\) that turns anticlockwise at the rate \(\omega\). At time \(t\) the phasor makes the angle \(\omega t + \phi\) with the horizontal, and its projection on the vertical axis is \(y\).
The three numbers of a sinusoid
The amplitude \(A\) is the length of the arm. The height of the tip can never be more than the length of the arm. So \(y\) swings between \(-A\) and \(+A\), and nothing else in the formula can change that range.
The angular frequency \(\omega\) is the rate of turning. One full turn is \(2\pi\) radians, and the time for one full turn is the period \(T\). So \(\omega T = 2\pi\), which gives \(\omega = 2\pi/T\). The frequency \(f = 1/T\) counts the turns in one second, so \(\omega = 2\pi f\). The mains supply in India has \(f = 50\,\text{Hz}\), so its \(\omega = 100\pi \approx 314\,\text{rad/s}\).
The phase is the angle of the arm at time \(t\), that is \(\omega t + \phi\). The constant \(\phi\) is the initial phase, the angle of the arm at \(t = 0\). Two sinusoids with the same \(A\) and \(\omega\) but different \(\phi\) have the same curve. One curve is only shifted in time relative to the other.
\[ \theta = \omega t + \phi \]
Why. the arm turns at a steady rate \(\omega\), starting from the angle \(\phi\).
\[ y = A\sin\theta \]
Why. the height of the tip is the side opposite \(\theta\) in a right triangle whose hypotenuse is \(A\).
\[ \theta(t + T) = \theta(t) + \omega T \]
Why. in a further time \(T\) the arm turns through a further angle \(\omega T\).
\[ \omega T = 2\pi \]
Why. the height repeats for the first time when the arm has made one full turn, \(2\pi\) radians, and is back where it started.
\[ T = \dfrac{2\pi}{\omega}, \qquad f = \dfrac{1}{T} = \dfrac{\omega}{2\pi} \]
Read it. solve step 4 for \(T\). The frequency is the number of turns in one second.
An alternating voltage is \(v = 5\sin(100\pi t + \pi/6)\) volts, with \(t\) in seconds. What is its frequency?
An arm of length \(6\,\text{cm}\) turns anticlockwise once every \(3\,\text{s}\), starting along the horizontal at \(t = 0\). What is the height of its tip at \(t = 0.25\,\text{s}\), in cm?
Ahead and behind
In the last chip of Fig 1.1 the arm starts at \(90°\), pointing straight up. The height starts at \(A\), and the curve is \(A\cos\omega t\). So a cosine is a sine whose arm starts a quarter turn ahead: \(\cos\omega t = \sin(\omega t + \pi/2)\). In the same way, \(-\sin\omega t\) is the arm half a turn ahead, and \(-\cos\omega t\) is the arm a quarter turn behind.
Take two sinusoids with the same \(\omega\). Their arms turn together, so the angle between the arms never changes. That fixed angle is the phase difference \(\Delta\phi\). The sinusoid whose arm is ahead, in the anticlockwise sense, reaches every peak first. It is said to lead, and the other to lag. The lead in time is the angle divided by the rate of turning, \(\Delta t = \Delta\phi/\omega\).
Two currents in a circuit are \(i_1 = 3\sin\omega t\) and \(i_2 = 3\cos\omega t\). Which statement is correct?
Time as an angle
The height of the tip changes unevenly, but the angle of the arm changes evenly. This is the whole trick. To find the time between two values of a sinusoid, find the two angles of the arm, subtract them, and divide by \(\omega\). One whole turn, \(\Delta\theta = 2\pi\), takes one period \(T\).
Now read the heights as angles. At the start, \(\sin\theta_1 = -2/4 = -1/2\), so \(\theta_1 = -30°\). At the end, \(\sin\theta_2 = 3.46/4 = 0.866\), so \(\theta_2 = 60°\). The block moves towards \(+x\) the whole way, so its height is rising. A rising height means the tip is on the right half of the circle, which fixes both angles. The arm turns \(60° - (-30°) = 90°\), a quarter turn. The time is \(T/4 = 0.6\,\text{s}\).
One value of the sine belongs to two angles. For example, \(\sin\theta = 1/2\) at \(30°\) and also at \(150°\). The height alone cannot tell you which angle is meant. The direction of motion decides. A rising height puts the tip on the right half of the circle, and a falling height on the left half. The arm also describes only a pure sinusoid with one steady \(\omega\). A damped swing, whose amplitude shrinks, has no arm of fixed length.
This is the worked example of Kinematics of SHM, lesson 5, done with the tool. 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?
Draw the phasor of the motion, \(x = 4\sin\theta\) in cm. The particle moves towards \(+x\), so \(x\) is rising and the tip of the arm is on the right half of the circle.
\(\sin\theta_1 = -2/4 = -1/2\), so \(\theta_1 = -30°\). \(\sin\theta_2 = 2\sqrt{3}/4 = \sqrt{3}/2\), so \(\theta_2 = 60°\).
The arm turns \(60° - (-30°) = 90°\). So \(\Delta t = (90°/360°) \times 2.4\,\text{s} = 0.6\,\text{s}\).
The particle takes \(0.6\,\text{s}\), a quarter of the period, for a stretch of \(5.46\,\text{cm}\). The time is short because the stretch contains the mean position, where the particle moves fastest.
The mains voltage is \(v = 311\sin(100\pi t)\) volts. How long after \(t = 0\) does the voltage first reach \(155.5\,\text{V}\)? Give the answer in milliseconds.
A particle in SHM of period \(1.2\,\text{s}\) leaves its mean position. How long does it take to reach half of its amplitude?
Why can the time between two positions of an oscillating body not be found as the distance divided by the speed? How does the turning arm give the time instead?
The speed of the body changes at every instant, so no single speed can be divided into the distance. The body's position is the height of the tip of an arm that turns at a steady rate \(\omega\). The angle of that arm grows evenly with time, even though the height does not. So we read the two positions as two angles, subtract them, and divide by \(\omega\). The direction of motion tells us which of the two possible angles to take.
The rotating arm is the reference circle of simple harmonic motion. The chapter Kinematics of SHM shows that the shadow of a point going round a circle moves in SHM. It also times motions by the angle swept. Superposition, Damped and Forced Oscillations compares two SHMs by the angle between their arms. Alternating Current and the LCR Circuit draws every voltage and current of the mains as such an arm, turning fifty times a second.
- Reading \(\omega\) as the frequency.\(\omega\) is the angle turned per second, in rad/s. The frequency is \(f = \omega/2\pi\); \(100\pi\,\text{rad/s}\) is \(50\,\text{Hz}\).
- Taking the wrong one of two angles that have the same sine.Use the direction of motion. A rising value puts the tip on the right half of the circle, and a falling value on the left half.
- Mixing degrees and radians in \(\omega t + \phi\).With \(\omega\) in rad/s, \(\omega t\) is in radians. Write the initial phase in radians too, or convert everything to degrees.