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9.2: Energy Reasoning

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    24136
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    The analysis of sound propagation in Section 9.1.3 required estimating the spring forces. Often, however, the forces or their effects are harder to track than are the energies. Then, as you’ll see in the next examples, we track the energy and look for the energy signature of a spring: the quadratic dependence of energy on displacement.

    9.2.1 Oscillation frequency of a spring–mass system

    To illustrate the energy method, let’s practice on the most familiar spring system by finding its natural (angular) oscillation frequency \(\omega_{0}\). The method requires finding its kinetic and potential energies. Because these energies vary in complicated ways, we use typical or characteristic energies.

    clipboard_e8f2d012a29b8b13748cc6dbd0d652a8f.png

    In terms of the amplitude A0, the typical potential energy is comparable to \(kA_{0}^{2}\). The typical kinetic energy is comparable to \(mv^{2}\), where \(v\) is the typical speed of the oscillating mass. This speed is comparable to \(A_{0}\omega_{0}\), because the mass travels a distance comparable to A0 in the characteristic time \(1/\omega_{0}\) (which corresponds to 1 radian, or approximately one-sixth of an oscillation period). Thus, the typical kinetic energy is comparable to \(mA_{0} \omega _{0}^{2}\).

    In spring motion, kinetic and potential energy interconvert, so the ratio

    \[\frac{\textrm{typical potential energy}}{\textrm{typical kinetic energy}}.\]

    should be comparable to 1. This bold conclusion is not limited to spring motion. For example, for gravitational orbits, the ratio, defined carefully using the time-averaged energies, is −2. More generally, the virial theorem says that, with a potential \(V \propto rn\), the energy ratio will be 2/n.

    Equating the typical energies gives an equation for \(\omega_{0}\):

    \[\underbrace{kA_{0}^{2}}_{E_{potential}} \sim \underbrace{mA_{0}^{2}\omega_{0}^{2}}_{E_{kinetic}}.\]

    The amplitude A0 divides out—another illustration that a spring’s period is independent of amplitude—giving \(\omega_{0} \sim k/m\). Because the energy ratio is 1 (due to the virial theorem), the missing dimensionless prefactor is 1.

    9.2.2 Vibrations of a piano string

    From springs to strings: A piano string is a steel wire stretched close to its breaking point—the high tension makes the string’s resistance to bending less important and the sound cleaner (as you investigated in Problem 9.17). When you push a piano key, a hammer bangs on the string and sets it into vibration—whose frequency we’ll estimate with a spring model.

    For a physical model, start with an unstretched piano string of length L. It is a bundle of springs and masses, so it acts like one large spring. Now stretch the string by putting it under tension T. Then hammer it.

    The hammer gives the atoms vertical velocities. Eventually the kinetic energy turns into potential energy, and the string gets a sinusoidal shape with a wavelength \(\lambda = 2L\) and a small amplitude y0.

    clipboard_e0d0b6b6772e507de81f48dc80e8751c1.png

    As we did for the simple spring–mass system (Section 9.2.1), we’ll find the typical potential energy in the string and the typical kinetic energy in the motion of the string. The potential energy comes from the tension force: The curved string is longer than the equilibrium string, so the tension force has done work on the string by stretching it; the string stores that work as its potential energy. The work done is the force times the distance, so

    \[E_{potential} \sim T \times \textrm{extra length}.\]

    To estimate the extra length, lump a piece of the curved string as the hypotenuse of a right triangle with base \(\cancel{\lambda}\), where \(\cancel{\lambda} \equiv \lambda/2 \pi\). This base represents 1 radian of the sine-wave shape. In 1 radian, a sine wave attains almost its full height (sin 1 ≈ 0.84), so the height of the triangle is comparable to the amplitude y0. The triangle then has slope \(tan \: \theta \approx y_{0}/ \cancel{\lambda}\). Because \(y_{0} << \cancel{\lambda}\), the opening angle \(\theta\) is small; thus, \(tan \: \theta \approx \theta \: textrm{and} \: \theta \approx y_{0}/\cancel{\lambda}\).

    clipboard_eb68e3e4b6dce59a8a4b86d5a5af18719.png

    Using our lumping triangle, we’ll find the fractional change in length between the hypotenuse and the base. Fractional changes, being dimensionless, require less algebra and are more widely applicable than absolute changes. To find the fractional change, rescale the triangle so that the base has unit length; then it has height \(\theta\) and hypotenuse \(\sqrt{1 + \theta^{2}}\). Because \(\theta\) is small, the square root is, by the binomial theorem, approximately \(1 + \theta^{2}/2\). Thus, the fractional increase in length is comparable to \(\theta^{2}\), which is \(y_{0}^{2}/\cancel{\lambda}^{2}\).

    clipboard_e0bd24da9abdd2916c3aa43525603d384.png

    The fractional increase applies to the whole string, whose length was L, so

    \[\textrm{extra length} \sim L (\frac{y_{0}}{\cancel{\lambda}})^{2}.\]

    The work done in making this much stretch, which is the potential energy in the string, is T × extra length, where T is the tension, so

    \[E_{potential} \sim TL (\frac{y_{0}}{\cancel{\lambda}})^{2}.\]

    As befits a spring, even a giant one composed of individual bond springs, the potential energy is proportional to the square of the amplitude y0.

    Now let’s estimate the kinetic energy in the motion of the string. As the string vibrates at the so-far-unknown angular frequency \(\omega\), the pieces of the string move up and down with a typical speed \(\omega y_{0}\). Thus,

    \[E_{kinetic} \sim \textrm{mass} \times (\textrm{typical speed})^{2} \sim \underbrace{\rho b^{2}L}_{m} \times \underbrace{\omega^{2} y_{0}^{2}}_{\sim v^{2}},\]

    where \(\rho\) is the string's density and b is its diameter. The kinetic energy is also proportional to the squared amplitude \(y_{0}^{2}\)--the other energy signature of a spring. Equating the energies gives the equation for \(\omega\):

    \[\underbrace{\rho b^{2} \cancel{L} \omega^{2} \cancel{y^{2}_{0}}}_{\sim E_{kinetic}} \sim \underbrace{T \cancel{L} (\frac{\cancel{y_{0}}}{\cancel{\lambda}})^{2}.}_{\sim E_{potential}}\]

    The length L and the squared amplitude \(y_{0}^{2}\) cancel, leaving

    \[\omega = \frac{1}{\cancel{\lambda}} \sqrt{\frac{T}{\rho b^{2}}}.\]

    Despite the extensive use of lumping, this result turns out to be exact--as do many energy-based spring analyses. The circular frequency \(f = \omega/2 \pi\) has the same structure:

    \[f = \frac{1}{\lambda} \sqrt{\frac{T}{\rho b^{2}}}.\]

    The wave propagation speed is \(f \lambda\) (or \(\omega \cancel{\lambda}\)), which is just \(\sqrt{T/\rho b^{2}}\). Let’s check that this speed makes sense. As a first step, it can be rewritten as

    \[v = \sqrt{\frac{T/b^{2}}{\rho}}.\]

    Because the numerator \(T/b^{2}\) is the pressure (force per area) applied to the ends of the string to put it under the tension T, the speed of these transverse waves is \(\sqrt{\textrm{applied pressure}/\rho}\). (They are called transverse waves because the direction of vibration is perpendicular, or transverse, to the direction of travel.) The speed is analogous to the speed of sound that we found in Section 5.5.4, \(\sqrt{\textrm{pressure}/\rho}\), where the pressure was the gas pressure or the elastic modulus. So our speed makes good sense.

    How long is the middle-C string on a piano?

    We can find the length from the propagation speed and the frequency. The frequency of middle C is roughly 250 hertz. The propagation speed is

    \[v =\sqrt{\frac{\sigma}{\rho}},\]

    where \(\sigma\) is the stress \(T/b^{2}\). This stress is also \(\epsilon Y\), where \(\epsilon\) is the strain (the fractional change in length). Using \(\sigma = \epsilon Y\) and \(c_{s} = \sqrt{Y/ \rho}\),

    \[v = \sqrt{\frac{\epsilon Y}{\rho}} = \sqrt{\epsilon} \sqrt{\frac{Y}{\rho}} = \sqrt{\epsilon} c_{s}.\]

    In steel, cs ≈ 5 kilometers per second. For piano wire, which is made of high-strength carbon steel, the yield strain is roughly 0.01. However, the string is not stretched quite so far. To provide a margin of safety, the strain \(\epsilon\) is kept to roughly 3 x 10-3. Then the transverse-wave speed is

    \[v \approx \underbrace{0.06}_{\sqrt{\epsilon}} \times \underbrace{5 \times 10^{3} \textrm{m s}^{-1}}_{c_{s}} = 300 \textrm{m s}^{-1}.\]

    clipboard_ea1d03c5c8ff7d6e2b21d2c96f89ada21.png

    At \(f \approx 250\) hertz, the wavelength is roughly 1.2 meters:

    \[\lambda = \frac{v}{f} \approx \frac{300 ms^{-1}}{250 Hz} = 1.2 m.\]

    The wavelength of this lowest, fundamental frequency is twice the string’s length, so the string should be 0.6 meters long. To check, I looked into our piano: 0.6 meters is almost exactly the length of the strings set into motion when I play middle C.

    9.2.3 Musical notes from bending beams

    Another musical device that we can model is a wooden or metal slat in a marimba or xylophone. Using spring models, proportional reasoning, and dimensional analysis, we’ll find how the frequency of the slat’s musical note depends on its dimensions.

    clipboard_e9f8ff7499e4009294b009ef11bc78ecd.png

    Thus, imagine a thin block of wood of length l, width w, and thickness h. It is supported at the two dots (or held at one of them) and tapped in the center. As it vibrates, its shape varies from bent to straight and back to bent. Here are the shapes shown in side view

    clipboard_ef0f151ccff32d05d9f4e8823abccb24b.png

    How does the slat’s width \(w\) affect the frequency?

    The answer comes from that cheapest kind of experiment, a thought experiment. Using a type of argument developed by Galileo in his study of free fall, lay two identical slats of width w side by side. Tapping both slats simultaneously produces the same motion as does tapping a slat of width 2w` (the result of gluing the two slats along the long, thin edge). So the width cannot affect the frequency.

    How does the slat’s thickness \(h\) affect the frequency?

    The slat, made of atoms connected by bond springs, acts like a giant spring–mass system. When the slat is straight (the equilibrium position), it has zero potential energy. The energy increases upon bending the slat, which stretches or compresses the bond springs. Thus, the slat resists bending. As befits a giant spring, its resistance to bending can be represented by a stiffness or spring constant k. (Mechanical engineers define a related quantity called the bending stiffness or the flexural rigidity, which has dimensions of energy times length. Our stiffness is an actual spring constant, with dimensions of force per length.)

    clipboard_eb5729bafdbf481c0fc682fa4510515f8.png

    Then, as befits a giant spring-mass system, the slat has a vibration frequency comparable to \(\sqrt{k/m}\), where m is its mass. Deciding how the thickness affects the frequency has split into two smaller problems: how the thickness affects the mass and how the thickness affects the stiffness. The first decision is not difficult: The mass is proportional to the thickness.

    How does the stiffness \(k\) depend on thickness?

    To answer this proportional-reasoning question, we’ll perform the thought experiment of bending each slat by the same vertical deflection y.

    clipboard_e17666f85d3f5171136e84be3a899ed98.png

    The stiffness k is proportional to the force F required to bend the beam (F = −ky). However, we won’t try to understand k by finding how the thickness affects F itself. Force is a vector, so finding the required force requires carefully bookkeeping many little forces and their directions to know which contributions cancel. Instead, we’ll find how the thickness affects the stored energy (the potential energy). As a positive scalar quantity, potential energy has no direction or even sign and is therefore easier to bookkeep.

    Because a slat is a big spring, the energy required to produce the vertical deflection y is given by

    \[E \sim ky^{2}.\]

    Because y is the same for the slats (an easy condition to enforce in a thought experiment), the energy relation becomes the proportionality \(E \propto k\). To find how k depends on thickness, let’s redraw the bent slats with a dotted line showing the neutral line (the line without compression or extension):

    clipboard_e04f006ee1bda2e67f1eedaf06a887709.png

    Above the neutral line, the bond springs along the length of the block get extended; below the neutral line, they get compressed. The compression or extension \(\Delta l\) determines the energy stored in each bond spring. Then the stored energy in the whole slat is

    \[E \sim E_{\textrm{typical spring}} \times N_{\textrm{springs}}.\]

    Because \(E \propto k\),

    \[k \propto E_{\textrm{typical spring}} \times N_{\textrm{springs}}.\]

    To find how Etypical spring depends on the slat’s thickness h, break the energy into factors (divide and conquer):

    \[E_{\textrm{typical spring}} \sim k_{\textrm{bond}} \times (\Delta l)^{2}_{\textrm{typical spring}}.\]

    Because the bonds in the two blocks are the same (the blocks differ only in thickness), this relation becomes the proportionality

    \[E_{\textrm{typical spring}} \propto (\Delta l)^{2}_{\textrm{typical spring}}.\]

    Therefore, the stiffness is

    \[k \propto (\Delta l)^{2}_{\textrm{typical spring}} \times N_{\textrm{springs}}.\]

    To find how \((\Delta l)_{\textrm{typical spring}}\) depends on h, compare typical bond springs in the thick and thin blocks—for example, a spring halfway from the neutral line to the top surface. Because the thick block is twice as thick as the thin block, this spring is twice as far from the neutral line in absolute distance.

    The extension is proportional to the distance from the neutral line—as you can guess by observing that it is the simplest scaling relationship that predicts zero extension at the neutral line (or try Problem 9.1). In symbols,

    \[(\Delta l)_{\textrm{typical spring}} \propto h.\]

    Furthermore, Nsprings is also proportional to h. Therefore, \(k \propto h^{3}\):

    \[k \propto (\Delta l)^{2}_{\textrm{typical spring}} \times N_{\textrm{springs}} \propto h^{3}.\]

    Doubling the thickness multiplies the stiffness by eight! The vibration frequency of a slat, considered as a giant spring, is

    \[\omega \sim \sqrt{\frac{k}{m}}.\]

    The mass is proportional to h, so \(\omega \propto h\):

    \[\omega \propto \sqrt{\frac{h^{3}}{h}} = h.\]

    Doubling the thickness should double the frequency. To test this prediction, I tapped two pine slats having these dimensions:

    \[ \underbrace{30 \textrm{ cm}}_{l} \times \underbrace{5 \textrm{ cm}}_{w} \times \left\{ \begin{array}{ll} 1 \textrm{ cm} \: \: (h \textrm{ for the thin slat}); \\ 2 \textrm{cm} \: \: (h \textrm{ for the thick slat}) \end{array} \right. \]

    To measure the frequencies, I matched each slat’s note to a note on a piano. The thin slat sounded like C one octave above middle C. The thick slat sounded like A in the octave above the thin block. The interval between the two notes is almost an octave or a factor of 2 in frequency.

    Now let’s extend the analysis to the xylophone slats, which vary not in thickness but in length. This extension bring us to the third scaling question.

    How does the slat’s length \(l\) affect the frequency?

    We already found the scaling relation between \(\omega\) (frequency) and w (width), namely \(\omega \propto w^{0}\); and between \(\omega\) and h (thickness), namely \(\omega \propto h\). By adding the constraints of dimensional analysis to these scaling relations, we can find the scaling relation between \(\omega\) and l.

    The quantities relevant to the frequency \(\omega\) are the speed of sound cs and two of the three dimensions: thickness ℎ and length l. The third dimension, the width, is not on the list, because the frequency, we already found, is independent of the width. (Instead of the speed of sound, the list could include the Young’s modulus Y and the density \(\rho\). As the only two variables containing mass, Y and cs would end up combining anyway to make cs.)

    These four quantities, made from two dimensions, make two independent dimensionless groups. The first group should be proportional to the goal \(\omega\). Because \(\omega\) and cs are the only quantities containing time, each as T−1, the group has to contain \(\omega/c_{s}\). Making \(\omega/c_{s}\) dimensionless requires multiplying by a length. Either of the two lengths works. Let’s choose l. (For the alternative, try Problem 9.2). Then the group is \(\omega l/c_{s}\).

    clipboard_e35a8e4d781e77cbb0b3c18be93a41d79.png

    The other group should not contain the goal \(\omega\). Then the only choices are powers of the aspect ratio \(h/l\). If we use \(h/l\) itself, the most general dimensionless statement is

    \[\frac{\omega l}{c_{s}} = f(\frac{h}{l}).\]

    The thickness scaling relation, \(\omega \propto h\), determines the form of \(f\), giving

    \[\frac{\omega l}{c_{s}} \sim \frac{h}{l}.\]

    Solving for the frequency,

    \[\omega \sim \frac{c_{s}h}{l^{2}}.\]

    As a scaling relation, \(\omega \propto l^{-2}\). Let's check the scaling exponent against experimental data. When my older daughter was small, she got a toy xylophone from her uncle. Its (metal) slats have the tabulated dimensions and frequencies. The lower and higher C notes (C and C') are a factor of 2 apart in frequency. If the scaling relation is correct, C should come from the longer slat, and the ratio of slat lengths should be \(\sqrt{2}\). Indeed, the measured length ratio is almost exactly \(\sqrt{2} \approx 1.414\):

    \[\frac{12.2 \textrm{ cm}}{8.6 \textrm{ cm}} \approx 1.419.\]

    clipboard_e3fe374377f2aded5727e187717786d38.png

    Exercise \(\PageIndex{1}\): Spring extension versus distance from the neutral line

    Consider a bent slat as an arc of a circle, and thereby explain why the spring extension is proportional to the distance from the neutral line.

    Exercise \(\PageIndex{2}\): Alternative dimensionless group

    Repeat the dimensional analysis for the dependence of the oscillation frequency on slat length but using \(\omega h/c_{s}\) and ℎ/l as the two independent dimensionless groups. Do you still conclude that \(\omega \propto l^{-2}\)?

    Solving the beam differential equation gives the vibration frequency as

    \[f \approx \frac{3.56}{\sqrt{12}} \times c_{s} \frac{h}{l^{2}}.\]

    The dimensionless prefactor \(3.56/\sqrt{12}\) is almost exactly 1. This example is one of the rare cases where using circular frequency (\(f\)) rather than angular frequency (\(\omega\)) makes the prefactor closer to 1.

    For my block of pine, a light wood, \(\rho \approx 0.5 \rho_{water}\) and \(Y \approx 10^{10}\) pascals, so

    \[c_{s} = \sqrt{\frac{Y}{\rho}} \sim \sqrt{\frac{10^{10} Pa}{0.5 \times 10^{3} \textrm{kg m}^{-3}}} \approx 4.5 \textrm{km s}^{-1}.\]

    For the thin block, ℎ = 1 centimeter and l = 30 centimeters, so

    \[f \approx c_{s} \frac{h}{l^{2}} 4.5 \times 10^{3} \textrm{m s}^{-1} \times \frac{10^{-2}m}{10^{-1}m^{2}} \sim 450 Hz.\]

    This estimate is reasonably accurate. The thin block’s note was approximately one octave above middle C, with a frequency of roughly 520 hertz.

    Exercise \(\PageIndex{3}\): Graphing the data on frequency versus length

    Check the scaling \(\omega \propto l^{-2}\) by plotting the xylophone data for frequency versus length on log–log axes. What slope should the graph have?

    Exercise \(\PageIndex{4}\): Finding the stiffness, then the frequency

    Use dimensional analysis to write the most general dimensionless statement connecting stiffness k to a slat’s Young’s modulus Y, width w, length l, and thickness h. What is the scaling exponent q in \(k \propto 2^{q}\)? Use that scaling relation and \(k \propto h^{3}\) to find the missing exponents in \(k \sim Y^{p}w^{q}l^{r}h^{3}\).

    Exercise \(\PageIndex{5}\): Xylophone notes

    If you double the width, thickness, and length of a xylophone slat, what happens to the frequency of the note that it makes?

    Exercise \(\PageIndex{6}\): Location of the node

    Here’s how you can predict the location of the node (where to hold the wood block) using conservation. Because the bar vibrates freely without an external force, the center of mass (the dot) stays fixed. Approximate the bar’s shape as a shallow parabola, find the center of mass (CM), and therefore find the node locations (as a fraction of the bar’s length). My daughter’s longest xylophone slat is 12.2 centimeters long with holes 2.7 centimeters from the ends. Is that fraction consistent with your prediction?

    clipboard_e50ad3208c9e8c48baee958740967818b.png


    This page titled 9.2: Energy Reasoning is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by Sanjoy Mahajan (MIT OpenCourseWare) .

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