The acceleration-displacement line
The law of SHM, \(a = -\omega^2 x\), is the equation of a straight line. Plot the acceleration \(a\) against the displacement \(x\). The graph passes through the origin, because \(a = 0\) at the mean position. Its slope is \(-\omega^2\), always negative. The line runs only from \(x = -A\) to \(x = +A\), because the particle never goes beyond the extremes. Fig 2.2 showed a particle riding such a line.
The line is a quick test for SHM. If a graph of \(a\) against \(x\) is straight, passes through the origin and slopes downwards, the motion is simple harmonic. Its slope then gives \(\omega\), and so the period. A curved graph, or one that slopes upwards, belongs to some other motion.
The graph of acceleration against displacement for a particle is a straight line through the origin with slope \(-9\,\text{s}^{-2}\). What is the angular frequency of the motion?
Solving the law of SHM
Since \(a = d^2x/dt^2\), the law of SHM is an equation for the position as a function of time.
\[ \frac{d^2x}{dt^2} = -\omega^2 x \]
Such an equation, which contains a derivative of the unknown function, is a differential equation. We need a function whose second derivative is \(-\omega^2\) times itself. Lesson 1 found two: \(\sin\omega t\) and \(\cos\omega t\). The derivation below builds every solution from them.
\[ \dfrac{d^2}{dt^2}\sin\omega t = -\omega^2\sin\omega t, \qquad \dfrac{d^2}{dt^2}\cos\omega t = -\omega^2\cos\omega t \]
Why. by the chain rule of Lesson 1, each differentiation brings a factor \(\omega\), and two of them bring a minus sign. Both functions satisfy the law.
\[ x = C\sin\omega t + D\cos\omega t \]
Why. the second derivative of a sum is the sum of the second derivatives. So any constant multiples of the two solutions, added, also give \(-\omega^2 x\).
\[ x = A\sin(\omega t + \varphi), \quad A = \sqrt{C^2 + D^2}, \quad \tan\varphi = \dfrac{D}{C} \]
Why. Lesson 1 showed that a sine plus a cosine is a single sine. The constants \(C\) and \(D\) become an amplitude and a phase.
\[ \text{two constants, } A \text{ and } \varphi \]
Read it. the law fixes only \(\omega\). The particle may start anywhere with any velocity, and these two starting facts fix \(A\) and \(\varphi\).
Every motion that obeys \(d^2x/dt^2 = -\omega^2 x\) can be written as \(x = A\sin(\omega t + \varphi)\). The angular frequency \(\omega\) comes from the law. The amplitude \(A\) and the initial phase \(\varphi\) come from the position and the velocity at \(t = 0\).
Sine or cosine?
Many books write SHM with a cosine, \(x = A\cos\omega t\). This book uses the sine. The two describe the same motion. By Lesson 1, \(A\cos\omega t = A\sin(\omega t + \pi/2)\). So a cosine is simply a sine with the initial phase \(\pi/2\): the particle starts at \(+A\) instead of at the mean position. Only the moment chosen as \(t = 0\) differs.
A particle moves as \(x = A\cos\omega t\). How does this motion compare with \(x = A\sin\omega t\)?
Finding \(A\) and \(\phi\) from the start
Suppose a particle in SHM starts at \(x_0\) with velocity \(v_0\). Try \(x = (v_0/\omega)\sin\omega t + x_0\cos\omega t\). At \(t = 0\) it gives \(x = x_0\). Its derivative at \(t = 0\) is \(v_0\). So this is the motion. By Lesson 1, its amplitude is \(\sqrt{x_0^2 + (v_0/\omega)^2}\).
A particle in SHM with \(\omega = 2\,\text{rad/s}\) is at \(x = 3\,\text{cm}\) at \(t = 0\), moving towards \(+x\) at \(8\,\text{cm/s}\). Find its amplitude and initial phase, and write its motion.
The motion is \(x = (v_0/\omega)\sin\omega t + x_0\cos\omega t = 4\sin 2t + 3\cos 2t\) cm.
The amplitude is \(A = \sqrt{4^2 + 3^2} = 5\,\text{cm}\), and \(\tan\varphi = 3/4\), so \(\varphi \approx 0.644\,\text{rad}\), about \(36.9°\).
So \(x = 5\sin(2t + 0.644)\) cm. Check: at \(t = 0\), \(x = 5 \times 0.6 = 3\,\text{cm}\).
The particle does not stop at \(3\,\text{cm}\). It is moving, so it goes on to \(5\,\text{cm}\) before it turns back.
Velocity and acceleration against time
Differentiate the displacement once to get the velocity, and again to get the acceleration. The chain rule gives a factor \(\omega\) each time.
\[ v = \frac{dx}{dt} = A\omega\cos(\omega t + \varphi), \qquad a = \frac{dv}{dt} = -A\omega^2\sin(\omega t + \varphi) = -\omega^2 x \]
The last step brings back the law of SHM, as it must. The velocity is largest in size, \(v_{\max} = A\omega\), where the cosine is \(\pm 1\). That happens where the sine is zero, at the mean position. The acceleration is largest in size, \(a_{\max} = A\omega^2\), where the sine is \(\pm 1\), at the extremes.
Read Fig 4.1 at a few instants. At \(t = 0\) the particle is at O, the velocity graph is at its peak, \(A\omega = 2\pi \approx 6.3\,\text{cm/s}\), and the acceleration is zero. At \(t = 1\,\text{s}\) the particle is at \(+4\,\text{cm}\). There the velocity is zero and the acceleration has its largest negative value, \(-A\omega^2 = -\pi^2 \approx -9.9\,\text{cm/s}^2\). The velocity graph reaches each peak a quarter period before the displacement graph does.
The phases make this exact. Write \(v = A\omega\sin(\omega t + \varphi + \pi/2)\): the velocity leads the displacement by \(\pi/2\). Write \(a = A\omega^2\sin(\omega t + \varphi + \pi)\): the acceleration leads by \(\pi\). A phase difference of \(\pi\) means the two are always opposite. When \(x\) is largest and positive, \(a\) is largest and negative.
A particle moves as \(x = 0.02\sin 50t\), with \(x\) in metres and \(t\) in seconds. What is its greatest acceleration, in m/s²?
For a particle in SHM, which statements are true?
Velocity against position
Often we know where the particle is but not when. Then we want the velocity as a function of position, with the time removed. The sine and the cosine of one phase are tied by \(\sin^2\Phi + \cos^2\Phi = 1\). That is enough. One step of the derivation below is left for you.
\[ x = A\sin\Phi \]
Why. the displacement, with the phase \(\Phi = \omega t + \varphi\).
\[ v = A\omega\cos\Phi \]
Why. the velocity found above.
?
Your turn. divide step 1 by \(A\) and step 2 by \(A\omega\), square both and add. Which line is it?
\[ v^2 = \omega^2(A^2 - x^2) \]
Read it. multiply through by \(A^2\omega^2\) and rearrange. The phase, and with it the time, has gone.
At a given \(x\), the particle has one speed but two possible velocities, one for each direction. At \(x = 0\) the speed is \(\omega A\), its greatest value. At \(x = \pm A\) it is zero. The equation \(x^2/A^2 + v^2/(A\omega)^2 = 1\) is the equation of an ellipse, with half-width \(A\) along the \(x\) axis and half-height \(A\omega\) along the \(v\) axis.
In Fig 4.2, with \(\omega = 2\,\text{rad/s}\), the ellipse reaches \(v = \pm 8\,\text{cm/s}\) at \(x = 0\). At \(x = 2\,\text{cm}\) the formula gives \(v = \pm 2\sqrt{16 - 4} \approx \pm 6.9\,\text{cm/s}\), as the readout shows. With \(\omega = 1\,\text{rad/s}\) the ellipse is half as tall. If the two axes were scaled so that \(A\omega\) looked as long as \(A\), the ellipse would become a circle. Lesson 5 finds that circle.
At which position is the speed of a particle in SHM greatest?
A particle performs SHM with amplitude \(5\,\text{cm}\) and angular frequency \(4\,\text{rad/s}\). Find its speed and its acceleration when it is \(3\,\text{cm}\) from the mean position, on the positive side. Compare them with their greatest values.
\(A = 5\,\text{cm}\), \(\omega = 4\,\text{rad/s}\), \(x = +3\,\text{cm}\). The time is not given, so use the equations in \(x\).
\(|v| = \omega\sqrt{A^2 - x^2}\) and \(a = -\omega^2 x\).
\(|v| = 4\sqrt{25 - 9} = 4 \times 4 = 16\,\text{cm/s}\). \(a = -16 \times 3 = -48\,\text{cm/s}^2\).
The greatest speed is \(A\omega = 20\,\text{cm/s}\) and the greatest acceleration is \(A\omega^2 = 80\,\text{cm/s}^2\). At \(3\,\text{cm}\) the particle has \(80\,\%\) of its top speed and \(60\,\%\) of its top acceleration.
A particle in SHM has a speed of \(12\,\text{cm/s}\) when it is \(3\,\text{cm}\) from the mean position, and \(9\,\text{cm/s}\) when it is \(4\,\text{cm}\) away. Find \(\omega\) and the amplitude.
Use \(v^2 = \omega^2(A^2 - x^2)\) at each position: \(144 = \omega^2(A^2 - 9)\) and \(81 = \omega^2(A^2 - 16)\).
Subtract the second from the first: \(63 = 7\omega^2\), so \(\omega^2 = 9\,\text{s}^{-2}\) and \(\omega = 3\,\text{rad/s}\).
Then \(A^2 - 16 = 81/9 = 9\), so \(A^2 = 25\) and \(A = 5\,\text{cm}\).
A particle performs SHM with amplitude \(10\,\text{cm}\) and period \(\pi\,\text{s}\). What is its speed when it is \(6\,\text{cm}\) from the mean position, in cm/s?
The needle of a sewing machine moves up and down very nearly in SHM. Suppose its stroke is \(3\,\text{cm}\), so \(A = 1.5\,\text{cm}\), and it makes \(1000\) stitches a minute. Then \(f \approx 16.7\,\text{Hz}\) and \(\omega \approx 105\,\text{rad/s}\). The needle is fastest, \(A\omega \approx 1.6\,\text{m/s}\), in the middle of its stroke, as it pierces the cloth. At the top and the bottom it stops, but its acceleration is \(A\omega^2 \approx 164\,\text{m/s}^2\), about \(16\) times \(g\). The machine is built to take these large accelerations at the ends of the stroke.
- Using \(A\omega\) for the greatest acceleration, or \(A\omega^2\) for the greatest speed.Each differentiation brings one \(\omega\): \(v_{\max} = A\omega\) and \(a_{\max} = A\omega^2\).
- Putting the speed and the acceleration at their greatest values in the same place.The speed is greatest at the mean position; the acceleration is greatest at the extremes.
- Measuring \(x\) in \(v = \omega\sqrt{A^2 - x^2}\) from an extreme.\(x\) is always the displacement from the mean position.
- Treating \(A\cos\omega t\) as a different motion from \(A\sin\omega t\).It is the same SHM with initial phase \(\pi/2\): it starts at \(+A\).