Corrected exercises: simple harmonic motion and mechanical waves (203-SN3-RE)

$\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{200}{0.50}}=20$ rad/s. Period: $T=\frac{2\pi}{\omega}\approx 0.31$ s; frequency: $f=\frac{1}{T}\approx 3.2$ Hz.

Released from rest at maximum displacement: $x(t)=A\cos(\omega t)=0.080\cos(20t)$ (in metres, $t$ in seconds). The cosine is the right choice because at $t=0$, $x=+A$ and $v=0$.

$v_{max}=A\omega=(0.080)(20)=1.6$ m/s, reached when passing through the EQUILIBRIUM position (where all the energy is kinetic). $a_{max}=A\omega^{2}=(0.080)(400)=32$ m/s², more than three times $g$, reached at the EXTREMES, where the spring pulls or pushes hardest and where, precisely, the speed is zero.

$v=\omega\sqrt{A^{2}-x^{2}}=20\sqrt{(0.080)^{2}-(0.040)^{2}}=20\sqrt{0.0048}\approx 1.39$ m/s. Energy: $E=\frac{1}{2}kA^{2}=\frac{1}{2}(200)(0.0064)=0.64$ J; at $x=0.040$ m, $E_{p}=\frac{1}{2}(200)(0.0016)=0.16$ J, so $E_{k}=0.48$ J and $v=\sqrt{\frac{2(0.48)}{0.50}}\approx 1.39$ m/s: the two routes agree. At half the amplitude, three quarters of the energy is still kinetic, not half: energy goes as the SQUARE of the displacement.

At equilibrium, the spring pulls as hard as the weight pushes: $kx_{eq}=mg$, so $x_{eq}=\frac{mg}{k}=\frac{(0.40)(9.8)}{25}\approx 0.157$ m, about 15.7 cm.

Let $X$ be the position measured from the NEW equilibrium position. The total extension is then $x_{eq}+X$, and the net force is $mg-k(x_{eq}+X)=mg-kx_{eq}-kX=-kX$, since $mg=kx_{eq}$. The weight has therefore disappeared from the equation entirely: it only SHIFTS the centre of the oscillations, without changing the restoring force. We recover $ma=-kX$, so $\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{25}{0.40}}\approx 7.91$ rad/s and $T=2\pi\sqrt{\frac{m}{k}}\approx 0.795$ s. An elegant check: since $\frac{m}{k}=\frac{x_{eq}}{g}$, we also have $T=2\pi\sqrt{\frac{x_{eq}}{g}}=2\pi\sqrt{\frac{0.157}{9.8}}\approx 0.795$ s. A vertical spring is therefore a pendulum whose length is its own static extension: measuring that extension with a ruler is enough to predict the period.

In PARALLEL, the two springs undergo the same extension and their forces add: $k_{eq}=k_{1}+k_{2}=50$ N/m, hence $T=2\pi\sqrt{\frac{0.40}{50}}\approx 0.562$ s, the initial period divided by $\sqrt{2}$. In SERIES, the two springs carry the same force (the weight, transmitted from one to the other) and their extensions add: $\frac{1}{k_{eq}}=\frac{1}{k_{1}}+\frac{1}{k_{2}}$, so $k_{eq}=12.5$ N/m and $T=2\pi\sqrt{\frac{0.40}{12.5}}\approx 1.124$ s, the initial period multiplied by $\sqrt{2}$. Note the kinship with capacitors: the spring constant adds in parallel, just as capacitance does, because it is the quantity that multiplies the extension in the force law $F=kx$.

Each half has a constant of 50 N/m, DOUBLE. Under a given force, the whole spring stretches by a certain amount, shared equally between its two halves: each half stretches by only half of the total while carrying the same force. Half the extension for the same force means twice as stiff. This is in fact the result of c) read backwards: the whole spring IS NOTHING BUT two half-springs in series.

$E=\frac{1}{2}kA^{2}=\frac{1}{2}(25)(0.050)^{2}\approx 3.13\times 10^{-2}$ J, and $v_{max}=A\omega=(0.050)(7.91)\approx 0.395$ m/s. Check: $\frac{1}{2}mv_{max}^{2}=\frac{1}{2}(0.40)(0.395)^{2}\approx 3.13\times 10^{-2}$ J. Gravity is absent from the balance for the same reason as in b): when the sum of gravitational and elastic energy is expanded around the new equilibrium position, the terms linear in $X$ cancel exactly and only $\frac{1}{2}kX^{2}$ remains, up to a constant that plays no role. The vertical spring is a horizontal spring in disguise, in energy as in dynamics.

$T=2\pi\sqrt{\frac{L}{g}}$ gives $L=g\left(\frac{T}{2\pi}\right)^{2}=9.8\times\frac{1}{\pi^{2}}\approx 0.99$ m. Almost exactly one metre: this is no historical accident, since one of the first definitions considered for the metre was precisely the length of the pendulum that beats seconds.

$T\propto\frac{1}{\sqrt{g}}$: $T_{Moon}=T\sqrt{6}=2.0\times 2.449\approx 4.9$ s. On the Moon, the same pendulum swings in slow motion, almost two and a half times slower.

For the pendulum, the restoring force IS the weight (its tangential component $mg\sin\theta$): the mass appears in the force AND in the inertia, and cancels, exactly as in free fall. For the spring, the restoring force $kx$ does not depend on the mass: inertia is the only thing at play and the mass survives in $T=2\pi\sqrt{\frac{m}{k}}$.

At altitude, $g$ decreases, so $T=2\pi\sqrt{\frac{L}{g}}$ increases: each swing lasts longer, and the clock LOSES time. To correct it, the pendulum must be shortened (raise the adjusting bob) to bring $T$ back to its value: a pendulum clock is, at heart, a gravimeter that does not know it.

By comparison with $A\cos(\omega t)$: $A=0.060$ m, $\omega=10$ rad/s, $T=\frac{2\pi}{10}\approx 0.63$ s, $f=\frac{1}{T}\approx 1.6$ Hz.

$v(t)=\frac{dx}{dt}=-0.60\sin(10t)$ m/s and $a(t)=\frac{dv}{dt}=-6.0\cos(10t)$ m/s². Maxima: $v_{max}=A\omega=0.60$ m/s and $a_{max}=A\omega^{2}=6.0$ m/s².

$a(t)=-6.0\cos(10t)=-(10)^{2}\times 0.060\cos(10t)=-\omega^{2}x(t)$. This relation says exactly that the restoring force is proportional and opposite to the displacement ($F=ma=-m\omega^{2}x$, Hooke's law): any system that satisfies it, whatever it is (spring, pendulum at small angles, LC circuit, vibrating molecule), oscillates sinusoidally. That is why it is taken as the definition: it is the common mechanical portrait of all harmonic oscillators.

$x=0$ when $\cos(10t)=0$, that is $10t=\frac{\pi}{2}$: $t=\frac{\pi}{20}\approx 0.16$ s, a quarter of a period after the start. Velocity: $v=-0.60\sin\left(\frac{\pi}{2}\right)=-0.60$ m/s, the maximum speed (in magnitude), directed toward negative $x$. Passing through equilibrium, all the energy is kinetic and the restoring force vanishes: the speed can only peak there.

$\mu=\frac{m}{L}=\frac{0.050}{2.0}=0.025$ kg/m. Speed: $v=\sqrt{\frac{F}{\mu}}=\sqrt{\frac{90}{0.025}}=\sqrt{3600}=60$ m/s.

$\lambda=\frac{v}{f}=\frac{60}{120}=0.50$ m.

The speed is a property of the MEDIUM: $v'=\sqrt{2}\,v\approx 85$ m/s. The frequency belongs to the SOURCE, and the vibrator has not changed: $f=120$ Hz still. The wavelength follows: $\lambda'=\frac{v'}{f}=\sqrt{2}\,\lambda\approx 0.71$ m. Remember the causal chain: the tension changes the medium, the medium changes $v$, and $\lambda$ adjusts; $f$ never changes along the way.

$k=\frac{2\pi}{\lambda}=\frac{2\pi}{0.50}\approx 12.6$ rad/m and $\omega=2\pi f\approx 754$ rad/s. Hence $y(x, t)=0.0040\sin(12.6x-754t)$ (SI units), the MINUS sign in front of $\omega t$ signalling propagation toward positive $x$. Check: $\frac{\omega}{k}=\frac{754}{12.6}\approx 60$ m/s, the speed recovered.

$\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{80}{0.80}}=10$ rad/s. All the initial energy is kinetic: $E=\frac{1}{2}mv_{max}^{2}=\frac{1}{2}(0.80)(0.50)^{2}=0.10$ J. At the extremes it is all potential: $\frac{1}{2}kA^{2}=E$ gives $A=\sqrt{\frac{2E}{k}}=\sqrt{\frac{0.20}{80}}=0.050$ m. (Or directly $A=\frac{v_{max}}{\omega}=\frac{0.50}{10}=0.050$ m: consistent.)

At $t=0$: $x=0$ and $v>0$, so a SINE is called for: $x(t)=0.050\sin(10t)$ (SI units).

$E_{k}=E_{p}$ means $E_{p}=\frac{E}{2}$: $\frac{1}{2}kx^{2}=\frac{1}{2}\left(\frac{1}{2}kA^{2}\right)$, hence $x^{2}=\frac{A^{2}}{2}$ and $x=\frac{A}{\sqrt{2}}\approx 0.035$ m, that is 3.5 cm: at about 71% of the amplitude, not at half.

$\sin(10t)=\frac{1}{\sqrt{2}}$ gives $10t=\frac{\pi}{4}$, so $t=\frac{\pi}{40}\approx 0.079$ s. With $T=\frac{2\pi}{10}\approx 0.63$ s, this is exactly $t=\frac{T}{8}$: the equal sharing falls at one eighth of a period, halfway (in TIME) between the passage through equilibrium and the extreme. The angle $\frac{\pi}{4}$, half of $\frac{\pi}{2}$, can be read directly on the reference circle: SHM rewards those who think in phase rather than in position. Each of the two energies oscillates TWICE per period of the motion, and their sum never moves.

Potential energy goes as the SQUARE of the position: at $x=\frac{A}{2}$, $E_{p}=\frac{1}{2}k\frac{A^{2}}{4}=\frac{E}{4}$ only, and $E_{k}=\frac{3E}{4}$. Half the way stores only a quarter of the energy: thinking “half the distance = half the energy” is the classic linear trap in a quadratic world.

For a rod hung from one end: $I=\frac{mL^{2}}{3}$ and $d=\frac{L}{2}$. So $T=2\pi\sqrt{\frac{mL^{2}/3}{mg(L/2)}}=2\pi\sqrt{\frac{2L}{3g}}=2\pi\sqrt{\frac{2(1.20)}{3(9.8)}}\approx 1.80$ s. The mass disappears because it appears both in the numerator (in $I$, the inertia that resists) and in the denominator (in $mgd$, the weight that restores): the two effects cancel exactly, just as for the simple pendulum and for free fall.

Simple pendulum of the same length: $T=2\pi\sqrt{\frac{L}{g}}=2\pi\sqrt{\frac{1.20}{9.8}}\approx 2.20$ s. The ROD is clearly faster, in the ratio $\sqrt{\frac{2}{3}}\approx 0.82$. The reason is that the mass of a rod is spread over its whole length, so on average much CLOSER to the axis than the bob of a simple pendulum, which is entirely at the end. Less inertia to rotate for a comparable restoring effect: the motion is livelier.

It would have to be placed at $L_{eq}=\frac{2L}{3}=0.80$ m from the axis: this is the length of the equivalent simple pendulum. This point is called the CENTRE OF OSCILLATION, or centre of percussion. Its best-known manifestation is the sweet spot of a baseball bat or a racquet: hitting the ball exactly there sends no jolt to the hands, because the impact produces no reaction at the axis of rotation. Hit elsewhere, and the hands absorb the vibration.

$I=\frac{1}{2}MR^{2}=\frac{1}{2}(0.50)(0.080)^{2}=1.6\times 10^{-3}$ kg·m². Period: $T=2\pi\sqrt{\frac{1.6\times 10^{-3}}{0.025}}=2\pi\sqrt{0.064}\approx 1.59$ s. The appeal of the torsion pendulum is twofold: its period depends on neither the amplitude (the wire obeys an angular Hooke's law) nor gravity, and its torsion constant can be made tiny, so the period very long and the device extraordinarily sensitive to the weakest torques. That is what allowed Cavendish to measure the gravitational attraction between two lead balls, and it is the same principle that, with a hairspring replacing the wire, makes the balance wheel of a mechanical watch beat at a rate that does not depend on the orientation of the wrist.

If the axis passes through the centre of mass, $d=0$: the weight no longer creates any restoring torque, the rod stays in neutral equilibrium in every position and does not oscillate at all ($T\to\infty$). In the general case, $T=2\pi\sqrt{\frac{\frac{L^{2}}{12}+d^{2}}{gd}}$. Minimizing $T$ amounts to minimizing $\frac{L^{2}}{12d}+d$, whose derivative $-\frac{L^{2}}{12d^{2}}+1$ vanishes for $d=\frac{L}{\sqrt{12}}=\frac{1.20}{3.464}\approx 0.346$ m. The period there is $2\pi\sqrt{\frac{0.12+0.12}{(9.8)(0.346)}}\approx 1.67$ s, shorter than that of the suspension by the end. There is therefore an optimal point of suspension, neither too close to the centre (no restoring torque left) nor too far (too much inertia): this is the compromise watchmakers tune.

$\omega_{0}=\sqrt{\frac{k}{m}}=\sqrt{\frac{200}{0.50}}=20$ rad/s. $\gamma=\frac{b}{2m}=\frac{0.80}{1.0}=0.80$ s$^{-1}$. $\omega'=\sqrt{400-0.64}=\sqrt{399.36}\approx 19.98$ rad/s: the rhythm has dropped by only 0.08%, eight hundredths of a percent. As in the RLC circuit, damping attacks first and almost exclusively the AMPLITUDE; it takes enormous friction to touch the period.

$e^{-\gamma t}=\frac{1}{2}$ gives $t=\frac{\ln 2}{\gamma}=\frac{0.693}{0.80}\approx 0.87$ s. During this time, with $T'=\frac{2\pi}{\omega'}\approx 0.314$ s, the block has completed $\frac{0.87}{0.314}\approx 2.8$ oscillations. In other words, fewer than three round trips are enough to lose half the amplitude, while the period has hardly moved.

$b_{c}=2\sqrt{km}=2\sqrt{(200)(0.50)}=2\sqrt{100}=20$ kg/s. Our $b=0.80$ kg/s is twenty-five times smaller: the system is very much UNDERDAMPED. The three regimes: underdamped ($b<b_{c}$, oscillations under a decaying envelope), critically damped ($b=b_{c}$, the fastest possible return to equilibrium without ever oscillating) and overdamped ($b>b_{c}$, slow return without oscillation). This is word for word the series RLC circuit of 203-SN2-RE, with $L$ in place of $m$, $\frac{1}{C}$ in place of $k$ and $R$ in place of $b$; the critical resistance $R_{c}=2\sqrt{\frac{L}{C}}$ is the exact transposition of $b_{c}=2\sqrt{km}$.

$A=\frac{F_{0}}{b\omega_{0}}=\frac{2.0}{(0.80)(20)}=0.125$ m, that is 12.5 cm. The same force applied CONSTANTLY would displace the block by only $\frac{F_{0}}{k}=\frac{2.0}{200}=0.010$ m, one centimetre. The amplification factor is therefore $\frac{0.125}{0.010}=12.5$, and we check that $Q=\frac{\omega_{0}m}{b}=\frac{(20)(0.50)}{0.80}=12.5$: at resonance, the oscillator amplifies exactly $Q$ times. A tiny force, but applied IN RHYTHM, produces an effect that the same force applied steadily never would. That is the whole mechanics of a child's swing.

A troop marching in step imposes a very regular periodic force, at a frequency close to that of some natural modes of a bridge; amplification by $Q$ can then turn modest pushes into destructive oscillations, hence the order to break step. The Tacoma Narrows bridge is a subtler case of the same family: the wind, shedding off the deck, sustained a torsional torque in phase with the motion, so that the energy input kept exceeding the little that damping dissipated, until failure. A car, on the other hand, is a case where resonance is the LAST thing you want: a poorly damped suspension would make the body bounce several times after each bump, and would amplify every road defect at the right frequency. The shock absorber is therefore tuned just below critical, enough for the body to return quickly to rest without oscillating, but not so much as to make the car uncomfortable. Resonance sought or avoided, it is always the same $Q$ being adjusted.

By comparison with $A\sin(kx-\omega t)$: $A=0.020$ m; $k=8\pi$ rad/m gives $\lambda=\frac{2\pi}{k}=0.25$ m; $\omega=200\pi$ rad/s gives $f=\frac{\omega}{2\pi}=100$ Hz; $v=\frac{\omega}{k}=\frac{200\pi}{8\pi}=25$ m/s; the minus sign in front of $\omega t$ indicates propagation toward POSITIVE $x$.

Each point of the string performs a vertical SHM of amplitude $A$ and angular frequency $\omega$: $v_{trans,max}=A\omega=(0.020)(200\pi)\approx 13$ m/s, to be compared with the 25 m/s of propagation. The two speeds measure different things: one is that of the MATTER (the point moves up and down without advancing), the other that of the PATTERN (the shape running along the string). One depends on the amplitude, the other does not: adding them or confusing them is THE trap of this chapter.

$v=\sqrt{\frac{F}{\mu}}$ gives $F=\mu v^{2}=(0.010)(25)^{2}=6.3$ N.

Same string, same tension: $v=25$ m/s unchanged (a property of the medium). The frequency doubles, $f=200$ Hz, so $\lambda$ is halved: $0.125$ m, and $k$ doubles: $16\pi$ rad/m; $\omega$ doubles: $400\pi$ rad/s. New equation (same amplitude): $y=0.020\sin(16\pi x-400\pi t)$. We check: $\frac{\omega}{k}=\frac{400\pi}{16\pi}=25$ m/s, still. The source dictates $f$, the medium dictates $v$, and $\lambda$ complies.