Vibrations on a String and Resonance - Experimental Physics

<strong>Vibrations</strong> on a String and ResonanceUmer Hassan and Muhammad Sabieh AnwarLUMS School of Science and EngineeringSeptember 7, 2010How does our radio tune into different channels? Can a music maestro shatter a crystal glass bybeating the tabla with a particular frequency and pitch? How does our ear distinguish betweentones in the multitude of sounds we hear every day? The answer to all of these questions liesin understanding the concept of resonance. The idea was discovered by Galileo Galilei with hisinvestigations of pendulums beginning in 1602. The present experiment gives you an introductionto the phenomenon of resonance, and the frequencies at which it occurs by visualizing the stationarywaves formed on a vibrating string. We hope you will enjoy this exercise of exciting and detectingstanding waves on a string.KEYWORDSTransverse wave · Longitudinal wave · Wave interference · Resonance · stationary waves · Circularmodes · normal modesAPPROXIMATE PERFORMANCE TIME 4 hours.1 Conceptual ObjectivesIn this experiment we will learn,1. the concept of wave,2. the difference between transverse and longitudinal waves,3. the phenomenon of wave interference,4. the concept of wave vector and wave number,5. derivation of the wave speed, and6. the formation of stationary waves.2 Experimental ObjectivesThe experimental objectives for the experiment include,1. exciting and detecting standing waves on a string,2. being able to identify where resonance occurs,3. distinguishing linear from nonlinear behaviors, and1

YDirection of PropagationvMotion of WaveParticleFigure 2: Illustration of particle motion w.r.t to wave propagation in a transverse wave.3.3 Sinusoidal or harmonic wavesThe sine wave is a mathematical function that describes a smooth repetitive oscillation. Mathematically,it can be expressed as,y(t) = Asin(ωt + ϕ), (1)where, A the amplitude, is the peak deviation of the function from its center position. ω, the angularfrequency, specifies how many oscillations occur in a unit time interval, in radians per second. It isgiven by ω = 2πf , f being the frequency of the wave. ϕ is the phase, specifying where in its cyclethe oscillation begins at t = 0. These concepts are illustrated in Figure 3.Oscillations dominate in real life All electromagnetic energy, including visible light, microwaves, radiowaves, and X-rays, can be represented by a sine wave or a combination of sine waves. At the lowestlevel, even matter oscillates like a wave. This is the realm of quantum physics. Other examplesinclude ocean waves, sound waves, and tides. Given the ubiquitous nature of waves, the currentexperiment sets to reveal some interesting properties. However, some more background theory isrequired before we can start our experiment.4 Wave Interference and Resonance4.1 Interference of wavesInterference is the phenomenon occurring when two waves meet while traveling along the samemedium. Constructive interference is a type of interference which occurs at any location alongthe medium where the two interfering waves have a displacement in the same direction. In otherwords, when the crest or trough of one wave passes through, or is super positioned upon, the crestor trough respectively of another wave, the waves constructively interfere. When waves interfere,amplitudes add. Figure 4 shows the two waves of different amplitudes constructively interfering togive a resulting wave of increased amplitude.Destructive interference is the type of interference which occurs at any location along the mediumwhere the two interfering waves have a displacement in the opposite direction. When the crest ofone wave passes through, or is super positioned upon, the trough of another wave, we say thatthe waves destructively interfere. We often say that when waves interfere, amplitudes add. Duringdestructive interference, since the positive amplitudes from one crest are added to the negativeamplitudes from the other trough, this addition can look like a subtraction. Refer to Figure 5 for ademonstration of this concept.If the phase difference is close to 180 ◦ , the resultant amplitude is nearly zero. When ϕ is exactly180 ◦ , the crest of one wave falls exactly on the valley of the other. The resultant amplitude is zero,corresponding to total destructive interference.3

(a)Amplitude1086420−2−4−6−8−100 2 4 6 8 1010864Angular frequency, w (rad/sec)(b)Amplitude20−2−4−6−8−100 1 2 3 4 5 61086Angular frequency, w (rad/sec)(c)Amplitude420−2−4−6−8−100 1 2 3 4 5 6Angular frequency, w (rad/sec)Figure 3: Illustration of variation of amplitude, frequency and phase of harmonic waves, (a) amplitudeof one wave is twice the other, (b) frequency of one wave is twice the other, and (c) one wavehas a phase difference of π 2with the other.⋆ Q 1. Two waves travel in the same direction and interfere. Both have the same wavelength,wave speed and an amplitude of 10 mm. There is a phase difference of 110 ◦ between them.(a) What is the resulting amplitude due to wave interference? (b) How much should the phasedifference change so that the resultant wave has an amplitude of 5 mm?4.2 Standing or stationary wavesStanding waves are formed by the interference of two harmonic waves of the same amplitudeand frequency (and therefore same wavelength), but traveling in opposite directions. Due to theinterference of the two waves, there are certain points called nodes at which the total wave is zeroat all times. The distance between two consecutive nodes is exactly half the wavelength. The points4

BoundariesAntinodeNodeλ\2Figure 6: Standing wave.particle of the free end. Since the rope and boundary are no longer attached and interconnected,they will slide past each other. So when a crest reaches the free end, the last particle of the wavereceives the same upward displacement; only now there is no adjoining particle to pull downwardupon the last particle of the wave to cause it to be inverted. The result is that the reflected wave isnot inverted. When an upward displaced wave is incident upon a free end, it returns as an upwarddisplaced wave after reflection and vive versa. Inversion is not observed in free end reflection, asshown in Figure 7.TransmittedWaveFreeBoundaryReflectedWaveFigure 7: Reflected wave from free end doesn’t gets inverted.4.3.2 Reflection from a fixed endIf a boundary is stationary i.e. not vibrating but fixed and a wave is traveling towards it, on reachingthe boundary, two things occur.1. A portion of the energy carried by the pulse is reflected and returns.2. A portion of the energy carried by the pulse is transmitted to the boundary, causing it tovibrate. But these vibrations are negligible, may contribute to sound or heat and are notdiscussed here.The reflected wave gets inverted. That is, if an upward displaced wave is incident towards a fixed endboundary, it will reflect and return as a downward displaced wave and vice versa, as shown in Figure 8.6

FixedBoundaryTransmittedWaveReflectedWaveFigure 8: Inversion take place from the fixed boundary.If there is continuous generation of transverse waves from one end and a transmitted wave getsinverted after being reflected from other end which is fixed, the wave interference took place andwe get a stationary wave as depicted in Figure 8. In our experiment we shall be forming standingwaves by continuously generating waves at one end and by keeping the other end fixed.4.4 ResonanceThe frequencies at which we get the stationary waves are the natural frequencies of the oscillatingsystem (in our case vibrating string). If we drive one end of the string and when the frequency ofthe driving force matches with the natural frequency of the string, standing wave is produced andthe string begins to move at large amplitude. This phenomenon is called as resonance.5 Wave Speed5.1 Basic definitionsThe wavelength λ is the distance between two consecutive crests or troughs in case of transversewave. The period T of the wave is the time required for any particular point on a wave to undergoone complete cycle of transverse or longitudinal motion. The frequency f is the number of thewave cycles completed in one second. The wave number k is the inverse of the wavelength and isexpressed as,Likewise, the angular frequency ω is,k = 2π λ . (2)ω = 2πT= 2πf . (3)The length L of the vibrating string in which standing waves are developed can be expressed asintegral multiples of half of the wavelength, i.e.,where n = 1,2,3,. . . is an integer. So, λ can be written as,L = nλ 2 , (4)λ = 2L n . (5)7

Substituting the value of λ into Equation 2, we get,k = nπ L . (6)The wave speed v can be written as,v = f λ = f 2π k = ω k . (7)5.2 Speed of wave on a string subject to tensionThe speed of the waves on a string depends upon the mass of the string element and the tensionT under which the string is stretched. The mass of the string element can be expressed in term ofthe linear mass density µ, which is the mass per unit length.vδlTTθRoFigure 9: A small portion of string under the action of forces.Let’s consider a small section of string of length δl as shown in Figure 9. Here v represents thewave speed and the direction in which the wave is traveling is identified by the arrow. The elementis part of an arc that is part of an an approximate circle of radius R. The mass δm of this elementis µδl. The tension T in the string is the tangential pull at each end of the segment. The horizontalcomponents cancel since they are equal and opposite to each other. However, each of the verticalcomponents is equal to T sinθ, so the the total vertical force is 2T sinθ. Since θ is very small, wecan approximate sinθ ≈ θ. Considering the triangle shown in the Figure 9, we find that, 2θ = δl \R.So, the net force F acting on the string element can be written as,F = 2T sinθ ≈ 2T θ = T δlR . (8)This is the force which is supplying the centripetal acceleration of the string particles towards O.The centripetal force F c acting on the mass δm = µδl moving in a circle of radius R with linearspeed v is,F c = δmv 2(9)REquating the two forces, we get,T δlR = δmv 2(10)R√Tv =µ . (11)8

Thus, the wave speed depends upon the tension and the linear mass density of the string.⋆ Q 2. Show that Equation 11 is dimensionally correct.⋆ Q 3. A transverse sinusoidal wave is generated at one end of long horizontal string by a barthat moves the end up and down through a distance of 1.5 cm.The motion is repeated at a rate of130 times per second. If a string has a linear density of 0.251 kg/m and is kept under a tension of100 N, find the amplitude, frequency, speed and wavelength of the wave motion.6 Experimental setupConsider a stretched string that is fixed from one end through a rigid support, is strung over a pulleyand a weight W is hung at the other end. The string can be set under vibrations using a mechanicaloscillator, which in our case is a speaker (woofer) fed with a signal generator. Let L be the lengthbetween the wedge and the oscillator as shown in Figure 10. If standing waves are establishedon the string, then the wave vectors can be written as,k n = nπ L, n = 1, 2, 3, . . . (12)LRigidSupportPulleyWedgeMechanicalVibratorWFigure 10: The experimental setup. A string is stretched between a rigid support and a mechanicalvibrator, which in our case, is a speaker. Tension in the string is introduced by a hanging mass.The relation between the angular frequency with the wave vector, is called the dispersion relationand depends upon the effective length and the tension. It can be derived as follows,Equating Equations 11 and 7, we obtain,√ωk = Tµ(13)Hence, for the bare string, the dispersion relation is given by,√ √Tω(k) =µ × k = n Tµ × π L . (14)The equation clearly shows that that the frequency modes ω(k) are directly proportional to thewave vector. Thus the dispersion relation for the string is linear.A normal mode of an oscillating system is a pattern of motion in which all parts of the systemmove sinusoidally with the same frequency and in phase. The frequencies of the normal modes areknown as its natural frequencies or resonant frequencies. A normal mode is characterized by a modenumber n, and is numbered according to the number of half waves in the vibrational pattern. If the9

vibrational pattern has one stationary wave, the mode number is 1, for two stationary waves, themode number is 2 and so on.Our goal in this experiment is to locate the normal modes of the vibrating string. We shall observethe formation of stationary waves on a vibrating string at resonance. The gist of the experimentis that the frequency ω will be varied until a pattern of standing waves is observed. When thiscondition is achieved, there will be an integer number of half-wavelengths formed on the string,which will be vibrating with a large amplitude. This is precisely what a normal mode is! We haveexcited a normal mode. The driving frequency is in resonance with a normal mode frequency. Wewill note down the frequency ω and the number n of half-wavelengths and will verify the relationshipgiven in Equation (14).6.1 Procedure⋆ Q 4. After receiving all the equipment, set up the apparatus according to the illustration inFigure 10.⋆ Q 5. Carry out the following experimental procedure to find the dispersion relation for thestring.1. Determine the value of µ of the string. (The density of the string is 8000 kg/m 3 ). What isthe uncertainty in the value?2. Place a driver at any place at the string such that the effective length is 1.5m.3. Attach 1 kg weight with the other end of the string.4. Now sweep the frequency slowly using the signal generator and find the frequencies at whichthe resonance occurs, i.e., where you observe the maximum amplitude standing waves. Startwith 1 Hz and increase the frequency with an increment of 0.1 Hz.5. Note down all the frequencies and plot a graph between frequency and the number of themode n. This is the desired dispersion relation.⋆ Q 6. Curve fit the data to Equation 14. What is your calculated value of T /µ?⋆ Q 7. Change the string tension using different weights (1.2 kg and 1.4 kg) and plot thedispersion relation for each case. Preferably, plot all your results with varying weight on thesame graph. Describe your observations.⋆ Q 8. Change the string length (1 m and 1.25 m) by changing the driver position and plot thedispersion relation for each case. Preferably, plot all your results with varying lengths on thesame graph. You should be able to describe your observations.6.2 Circular modes in the vibrating stringThe string is being driven in a vertical direction and therefore is expected to vibrate in the verticalplane only. At or near resonance the string has vertical as well as horizontal oscillations. This canbe easily visualized by having a side and top view of vibrating string. At this point each elementof the string is moving in a circle about the equilibrium position of the string. These oscillatorypatterns are called circular modes [3].⋆ Q 9. Is there any change in the string tension as the string vibrates?10

⋆ Q 10. Can you identify the reason for occurrence of these circular modes?⋆ Q 11. Observe what is the effect of decreasing the oscillator’s amplitude on the circularmodes?6.3 Resonance modes on a loaded string6.3.1 TheoryThe atoms in a crystalline structure are not at rest, they vibrates about their mean positions underthe influence of some energy field gradient. This vibrations of the atoms are called lattice vibrations.The energy present in the lattice vibrations can be looked as a series of superimposed sound or strainwaves whose frequency spectrum can be determined by the elastic properties of the crystal. Thequantum of energy of elastic wave is called phonon. The lattice vibrations can be visualized as asystem of identical atoms which are elastically coupled to each other by strings as shown in Figure11.These lattice vibrations can be simulated experimentally by uniformly loading a string with beads orsmall masses m with an interbead distance a, the resulting dispersion relation would no longer belinear. We will ’simulate‘ the behavior of the chain of the atoms by using a string loaded with rosarybeads.u n-1u n u n+1n-1 n n+1aFigure 11: Monoatomic lattice vibrations. A simplified picture of a linear chain of atoms.6.3.2 ExperimentA uniformly loaded string with beads can be used to simulate vibrations in a mono atomic crystallinelattice. Let the total number of beads be N which are uniformly loaded with an inter bead distance(lattice constant) ‘a’. Then the wave vector can be expressed as,where L = (N + 1)a.k n =nπ, n = 1, 2, 3, . . . , N. (15)(N + 1)a⋆ Q 12. Carry out the following experimental procedure to find the dispersion relation for thebare string.1. Find the mass of the beads.2. Measure the interbead distance.3. Place a driver at the string such that the effective length is 1.5m.4. Attach a 1 kg weight with the other end of the string.5. Now sweep the frequency slowly using signal generator and find the frequency modes at whichresonance occurs, i.e., where you observe the maximum amplitude standing waves.11

6. Note down all the frequencies and plot a graph between frequency and number of mode n.This is the desired dispersion relation.⋆ Q 13. Explain the non linearities in the dispersion relation and any other differences from thebare string.References[1] S. Parmley, Tom Zobrist, T. Clough, A. Perez-Miller, M. Makela, and R. Yu, “Vibrationalproperties of a loaded string”, Amer. J. Phys. 63, 547-553 (1995).[2] Resnick, Halliday, Krane, “Physics”, Chapter 19, John Wiley & Sons, 1992.[3] John A. Elliott, “Nonlinear resonance in vibrational strings”, Amer. J. Phys. 50, 1148-1150(1982).12