The socket in an Indian home is marked \(230\,\text{V}\), \(50\,\text{Hz}\). The voltage across its two holes is not steady. It rises to about \(325\,\text{V}\), falls through zero to \(-325\,\text{V}\) and returns, fifty times in every second. An electrical engineer writes the voltage as \(v = 325\sin(100\pi t)\), with \(t\) in seconds. Every number in that short equation means something. What is the voltage \(2.5\) milliseconds after it passes zero?
The sine and the cosine as graphs
On the unit circle, \(\sin\theta\) is the height of the tip of a radius turned through \(\theta\). Let the angle grow steadily and plot the height against the angle, in radians. The graph starts at 0 and rises to 1 at \(\theta = \pi/2\). It returns to 0 at \(\pi\), falls to \(-1\) at \(3\pi/2\) and is back at 0 at \(2\pi\). Then the radius begins a second turn, and the graph repeats.
An angle of \(\theta + 2\pi\) ends at the same point of the circle as \(\theta\). So \(\sin(\theta + 2\pi) = \sin\theta\) and \(\cos(\theta + 2\pi) = \cos\theta\) for every angle. A graph that repeats itself exactly after a fixed step is called periodic. The step, here \(2\pi\), is its period in the angle.
The cosine is the width of the same triangle instead of its height. Its graph is the sine graph moved back by a quarter turn. It starts at its top, \(\cos 0 = 1\), and it reaches each value a quarter turn before the sine does. So \(\cos\theta = \sin(\theta + \pi/2)\). The two curves are one shape, started at different points.
Time inside the sine
Physics puts time inside the sine. The angle becomes \(\omega t + \phi\), which grows steadily as time passes, like the angle of a turning radius. The quantity \(y = A\sin(\omega t + \phi)\) is called a sinusoid. Three constants fix it completely, and each has a name.
The first is the amplitude \(A\). The sine never goes above 1 or below \(-1\). So \(y\) swings between \(+A\) and \(-A\), and \(A\) is its largest value. The distance from the lowest point to the highest point, \(2A\), is called the peak-to-peak value.
The second is the angular frequency \(\omega\), in radians per second. The angle grows by \(\omega\) in each second. The curve repeats when the angle has grown by \(2\pi\). That takes a time \(T\) with \(\omega T = 2\pi\), so the period is \(T = 2\pi/\omega\). The frequency, the number of cycles in each second, is \(f = 1/T = \omega/2\pi\), in hertz.
The third constant is the phase \(\phi\). At \(t = 0\) the angle is \(\omega \cdot 0 + \phi = \phi\), so \(y(0) = A\sin\phi\). The phase says where in its cycle the sinusoid was when the clock was started. It is measured in radians, like the rest of the angle.
The voltage of the mains is \(v = 325\sin(100\pi t)\) volts, with \(t\) in seconds. Find its peak value, its angular frequency, its period and its frequency. Find the voltage at \(t = 2.5\,\text{ms}\), \(5\,\text{ms}\) and \(10\,\text{ms}\).
Compare with \(y = A\sin(\omega t + \phi)\). The peak value is \(A = 325\,\text{V}\), \(\omega = 100\pi = 314\,\text{rad/s}\) and \(\phi = 0\).
\(T = 2\pi/100\pi = 0.02\,\text{s} = 20\,\text{ms}\), and \(f = 1/T = 50\,\text{Hz}\).
At \(2.5\,\text{ms}\) the angle is \(100\pi \times 0.0025 = \pi/4\), so \(v = 325 \times 0.707 = 230\,\text{V}\).
At \(5\,\text{ms}\) the angle is \(\pi/2\) and \(v = 325\,\text{V}\), the peak. At \(10\,\text{ms}\) the angle is \(\pi\) and \(v = 0\).
The wall socket has given up its numbers. The \(325\) is the peak, the \(100\pi\) makes fifty cycles a second, and the missing phase means the clock started at a zero. Two and a half milliseconds later, an eighth of a cycle, the voltage is \(230\,\text{V}\).
A quantity varies as \(y = 5\sin(\pi t/2)\), with \(t\) in seconds. What is its period?
The mains voltage is \(v = 325\sin(100\pi t)\) volts. How many milliseconds after \(t = 0\) does the voltage first reach \(325\,\text{V}\)?
The phase slides the curve
Write the sinusoid as \(A\sin(\omega t + \phi) = A\sin\big(\omega(t + \phi/\omega)\big)\). This is the curve \(A\sin\omega t\), but with every event happening a time \(\phi/\omega\) earlier. So a positive phase slides the whole graph to the left, towards earlier times, by \(\phi/\omega\). The shape, the height and the period stay the same.
Some phases are worth knowing by sight. A phase of \(\pi/2\) slides the sine a quarter period earlier, and the result is the cosine: \(A\sin(\omega t + \pi/2) = A\cos\omega t\). A phase of \(\pi\) slides it half a period, and \(A\sin(\omega t + \pi) = -A\sin\omega t\). A phase of \(2\pi\) slides it a whole period, back onto itself.
Two sinusoids of the same \(\omega\) keep a fixed difference between their phases. If the difference is zero, the two rise and fall together; they are in phase. If it is \(\pi\), one is at its top when the other is at its bottom; they are in opposition. The one with the larger phase reaches each peak earlier, by the difference divided by \(\omega\). It is said to lead.
The shift in time is \(\phi/\omega\), not \(\phi\). A phase of \(\pi/2\) on the mains, \(\omega = 100\pi\), is a shift of only \(5\,\text{ms}\). The direction matters too. A plus sign inside the sine moves the graph to the left, earlier, not to the right. Finally, keep the whole angle in one unit. The term \(\omega t\) is in radians, so a phase of \(30°\) must be written as \(\pi/6\) before it is added.
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 \(y = A\sin(\omega t + \phi)\): \(A = 0.03\,\text{m}\), \(\omega = 4\pi\,\text{rad/s}\) and \(\phi = \pi/6\).
\(f = \omega/2\pi\), \(T = 1/f\) and \(x(0) = A\sin\phi\).
The amplitude is \(3\,\text{cm}\). \(f = 4\pi/2\pi = 2\,\text{Hz}\), so \(T = 0.5\,\text{s}\). At \(t = 0\), \(x = 0.03\sin(\pi/6) = 0.03 \times 0.5 = 0.015\,\text{m}\).
The particle starts \(1.5\,\text{cm}\) on the positive side. Its phase \(\pi/6\) lies before \(\pi/2\), so the sine is still rising, and the particle moves towards \(+x\).
Which sinusoid is the same function as \(A\cos\omega t\)?
In an AC circuit with a coil, the voltage is \(v = v_0\sin\omega t\) and the current is \(i = i_0\sin(\omega t - \pi/2)\). The supply is at \(50\,\text{Hz}\). By how many milliseconds does the current reach its peak after the voltage?
A quantity varies as \(y = A\sin(\omega t + \phi)\). Which statements are true?
Kinematics of SHM writes every simple harmonic motion as \(x = A\sin(\omega t + \phi)\) and reads its amplitude, period and initial phase. Superposition, Damped and Forced Oscillations compares two oscillations of different periods. Introduction to Waves describes a harmonic wave by \(y = A\sin(kx - \omega t)\), in which every particle of the string moves as a sinusoid. Propagation of Sound Waves finds the pressure a quarter cycle ahead of the displacement. Alternating Current and the LCR Circuit writes the mains voltage and current as sinusoids. The electromagnetic wave carries its electric and magnetic fields in step.
- Taking the phase \(\phi\) itself as the shift in time.The shift is \(\phi/\omega\). Divide the phase by the angular frequency.
- Sliding the graph to the right for a positive phase.\(A\sin(\omega t + \phi)\) reaches each value earlier, so the graph moves to the left.
- Writing \(T = \omega/2\pi\).That is the frequency. The period is \(T = 2\pi/\omega\), the time for the angle to grow by \(2\pi\).
- Calling the peak-to-peak swing the amplitude.The amplitude is half of it, the largest distance from the middle.