Vibration and Waves: Simple Harmonic Motion - Coursework Example

Vibration and waves Name Course Tutor Date Vibration and waves 2010 Section A 1 (a) Simple harmonic motion (SHM) is a type of periodic movement of an object that its acceleration is normally directed at a given equilibrium mark and is proportional to the displacement at that instant. From the definition of SHM, acceleration resulting to motion of the object is given by ‘a’, which is proportional and opposite of its displacement x to equilibrium point. i.e. a(t)α –x(t) If k is proportionality constant, then keeping in mind that acceleration is the second derivative of displacement, the differential equation, therefore, becomes x``(t) = -k x(t). (b) Transverse normal modes (c) A 0.025 kg block on a horizontal frictionless surface is attached to horizontal spring with a spring constant 0.4 m-1. the block is given both potential and kinetic energy by displacing it 0.1 m to the right of equilibrium and imparting a velocity of 0.4 m/s also to the right. Calculate: the amplitude of the motion, and the maximum acceleration of the block. i) Amplitude Mass = 0,025 Spring constant = 0.4 m, displacement 0.1 m and velocity = 0.4 m/s y= a sin(ωt) = a sin (k/mt)1/2 Velocity v = displacement/ time Implying that time t = displacement/ velocity t = 0.1/0.4 = 0.25 seconds From Hooke’s law f = ma or f =1/2ke2 Amplitude a = displacement/ sin (ωt) = 0.1/ sin (0.4/0.025*0.25)1/2 = 2.87 m ii) The force produce by the block f = 1/2ke2 = 1/2* 0.4* 0.12 = 0.002J. F= ma whereby maximum acceleration will be given force/mass= 0.002/0.025 = 8.0 *10-2 ms-2 d) Expression for time period of simple pendulum is given by T = 1/f, where f is the frequency of oscillation. i) When the length of the pendulum is double, it implies that the distant of displacement is increased, but we know that periodic motion of a pendulum is proportional to the displacement; therefore, the periodic time will reduce. ii) From the equation F = ma where F is the force of oscillation, m is the mass and ‘a’ is acceleration. Period time depends of the force produce by the oscillating object; thus when the bob mass is halved the force of oscillation will reduce by half as well. e) Consider two strings, both fixed at each end, one twice the length of the other. Do they have any lowed wavelengths for standing waves in common? Two strings of different lengths fixed at both ends have same lowed wavelengths for standing waves because the nodes and the antinodes will be at similar points. This is because the standing points appear to be equal in both cases and the wavelengths will only depended on the frequency of vibration of each string. f) The wave equation is given by V = fλ. This comes from the speed of a wave which is given by speed = wavelength/ period, and because frequency is the reciprocal of the period, the expression 1/f can be replaced to come up with speed = frequency * wavelength (V = fλ). This equation is the wave equation and it defines mathematically the association between velocity (v) of the wave, its frequency (f), and wavelength (λ). From the equation, we can see that the velocity of a wave is directly proportional to the frequency and its wavelength. g) The principle of superposition state that when two or more waves move over each other, the outcome disturbance at some point through the medium usually could be determined from displacement produced by each wave this is a case of all linear systems. h) Show that iodine is diatomic Iodine atomic mass = 127g mol-1 Temperature for iodine to vapor = 400K Nodes distance = 6.77 cm Frequency = 1000Hz, and γ = 1.4 Usually, the kinetic energy of rotation is Еrot = L2/2I, where L is the angular momentum and I is the molecule moment of inertia. Molecule translational energy is represented by the kinetic energy Е Trans = ½ mv2 where m-mass of molecule and v-velocity. Diatomic molecule has inertia given by I = µγ2o where γo is average distance between atoms for different molecules and µ is the molecule reduced mass. Thus, substituting the angular momentum and inertia moment into Еrot, diatomic rotational energy change to Еrot = l(l + 1)ћ2/2µγ2o l = 0, 1, 2, … Now replacing the values into the equation Еrot, 1 = ћ2/2MI2γ2o ≈ (1.05 * 10-34 J.s)/2(127*1037*1.4) = 3.1* 107J. =3.1* 107 J. Thus Iodine = 3.1* 107/(400*6.77*1000) ≈ 11.87 ≈ 12. Therefore, Iodine is a diatomic because in standards situations 12 is solid iodine and not vapor. i) If threshold level of hearing is Io = 10-16 Wcm-2, intensity level of a sound of intensity I is given by: B(decibels) = 160 + 10log I Io = 10-16 implying that log Io = -16 log 10 Where log Io = -16* log 10 Log I/log Io = e-16 + log 10 = 160 + 10log I Log I/log Io = decibel therefore B(decibel) = 160 + 10 log I. j) Stationary source frequency = 1000 Hz Sound travel with the velocity of 340 ms-1 From the wave equation V= fλ implying that λ = V/f =340/1000 = 3.4*10-1m With source moving at 30 ms-1, wavelength will be 30/1000) = 3.0*10-2m Section B 3 (i) showing that the superposition of two waves is a standing wave. In essence, waves combine together either constructively or destructively. When one physical interference of medium create the wave y1(x, t) and other given interference creates the wave y2(x, t), then both function equally, and the resultant wave is given by yres(x, t) = y1(x, t) + y2(x, t) Now taking two waves with equal velocity, amplitude, frequency, and direction of movement but different phase constant, we get y1 =ymsin(kx - ωt) and y2 = ymsin(kx-ωt + Φ) using trigonomeytry to add the two wave equations we get y`(x, t) = y1(x, t) + y2(x, t) = The resulting wave entails a new phase but largely essential is the new amplitude that is based on ym and Φ: Therefore, y’m = 2ymcos. At the time when Φ = 0, new amplitude will be 2ym, and waves are said entirely in phase and their sum forms a constructive wave with a larger wave being formed. However, Φ = п, the new amplitude is zero and waves are entirely out of phase. Their addition forms destructive waves or complete cancelation. Again adding y1(x, t) and y2(x, t) we get y’m(x, t) = y1(x, t) + y2(x, t) = This is not a moving wave because it is not of the type f(kx± ωt). This is sinusoidal function with coordinate x, as a multiple of moderating factor cos(ωt). Therefore, because equal spatial trend of vibration remains at fixed point, a wave of the above equation is thus a standing wave. For this wave, points with no displacement i.e. at sin(kx) = 0 implying that x = , where n is an integer. Nodes occur at this points of standing wave, whereas areas of maximum displacement i.e sin(kx) = ±1 where x = are referred to as antinodes of standing waves. Successive antinodes and nodes are split by λ/2, where λ is wavelength of initial waves creating the standing wave. A rope fixed on both ends is like a musical string, therefore vibration of the rope comprises of nodes at either side limiting the likely vibrations. In this case, we assume that the rope is of length L thus, have standing wave having a wavelength twice as long as the rope (wavelength λ = 2 L). The following series shows the different four normal modes of transverse waves on the rope. Images of the first four modes of rope vibration with fixed ends (ii) Frequencies of these modes can be derived as follows. For any wave, frequency is the ratio of wave speed to the length of wave i.e. f= v/ λ. In association with rope length L, the waves have lengths 2L, L, 2L/3, L/2. This can be given as 2L/n, for this case n is number of harmonic. The first mode or fundamental frequency is given by f1 = v/ λ1 = v/ 2L Second harmonic frequency is given by f2 = v/ λ2 = 2v/2L = 2f1 Third harmonic frequency f3 = v/ λ3 = 3v/2L = 3f1 Fourth harmonic frequency f4 = v/ λ4 = 4v/2L = 4f1, in general, the frequency of nth harmonic is given by fn = v/ λn = nv/2L = nf1 (iii) The sketch of the first, second, and third harmonics in a pipe closed at one end and open on the other. Open-Closed Waveforms fundamental frequency 1st harmonic fo 1st overtone 3rd harmonic f1 = 3fo 2nd overtone 5th harmonic f2 = 5fo For this case, the subscript of frequency resembles the order of overtone, and not the harmonic order. (iv) Electromagnetic waves between mirrors Helium-neon operating on wavelength 632.9924nm and mirror spacing 316.4962mm. This laser operates on longitudinal modes of operation where the Fabry-Perot resonator physical dimension inflicts an extra limitation on the outcome beam components. The next possible wavelength is given by W =, where w illustrates the likely wavelength of oscillation, L is the distance between mirrors, and n is larger integer Therefore W = = 6.677*10-4m. 5. (i) Boundary conditions applying to two stretched strings When the strings posses the equivalent impedance (Z2=Z1) then, ρ = When the string is heavy for example a wall, Z2= and. Ρ= The outcome would be same amplitude waves leading to opposite directions that are characteristics for standing waves. When ρ=-1, the reflected wave zeros the incident wave at junction resulting to node at junction of the strings. When the 2nd string is very light then Z2/Z1 G ρ = This condition leads to standing wave however, because reflected wave sums to incident wave at junction there would be antinode at junction. When Z1> Z2 (heavy to light string) reflection coefficient is positive and reflected wave entails equal orientation like incident wave. When Z1< Z2 (light to heavy string) reflection coefficient is negative and reflected wave is spanned equivalent to incident wave. (ii) Two strings at border condition, the transverse coordinate y and the derivative of transverse coordinate dy/dz should together be progressing. Taking the situation of incident on left border. Symbolize this wave by y1(z-vt). The wave shall be partly transmitted and partly reflected at the border. Assume the reflected wave is yR(z+vt) and the transmitted wave is yT(z-vt)…...1 Recalling z±vt shows the direction of spreading, then total wave in first string is, y1 = yT(z-vt) + yR(z+vt)……………………2 And total wave in second string is y2 = yT(z-vt)……………………3 Boundary conditions thus becomes yT(z=0, t) = + yR(z=0, t) = yT(z = 0, t)………4 …...5 These relations can be recast according to previous Eq. to become …………………6 The added different signs show the reflected waveform because of different direction of propagation. Further, the velocities are different. From Equ 5 -…………………7 Integrating this equation gives - yT(z=0,t)-yR(z=0,t)=…...8 However, integration should always be zero because when yT=0, yR and yT should, as well be zero. Thus, the tension in the two strings should be same when strings are not fixed between the strings Then Eq. 8 becomes yT(z=0,t)-yR(z=0,t)=…………9 Then using Eq. 5 and 9 yT(z=0,t)-yR(z=0,t)= …………………10 Therefore, we have resolved the reflection coefficient ρ that provides the fraction of incident wave front, which is reflected at border at some time. Equivalently, the transmission coefficient т could be derived from Equ. 5 and 10, yT(z=0,t) = yR(z=0, t) = yT(z=0, t) yT(z=0, t) + ρyT(z=0, t) = yT(z=0, t) = т ………………………….11 (iii) Consider a glass coated with a film. Explain why there is no reflection from surface if the film is of thickness λ/4. Coating a surface interface usually creates some optical coats on the surface of glasses or lenses that reduce reflection. This enhances the efficiency of system because light is mislaid. In most cases, the misled reflection creates contrast of image by removing some stray light. Furthermore, in some cases, coating films has alternating layers of divergent refractive index hence produce destructive interference of beams at reflected interfaces. Film of thickness λ/4.makes the interface more dense and able to absorb all the incident rays thus there will be no any reflection. It is therefore, advisable to specify a wavelength while designing or making coatings to surfaces. Vibration and waves 2009 Section A 1) (a) Simple harmonic motion (SHM) is a type of periodic movement of an object that its acceleration is normally directed at a given equilibrium mark and is proportional to the displacement at that instant. (b (c) Mass of block = 400 g Displacement = 25cm Period = 0.75s Velocity = 100 ms-1 Spring constant 0.4m-1 Period T = 1/f, implying that angular frequency (f) of the system would be f = 1/T = 1/0.75 = 1.333 Hz However, we know that force (F) producing the velocity is F = -kx Therefore force F = 0.4*0.25 = 0.10N Displacement y = A sinωt = Asin √(kt/m) And velocity ν = ωAcosωt Amplitude A of oscillation = displacement/sin(ωt) Velocity = displacement/time therefore, time to achieve that velocity = displacement/ velocity = 25/100 = 0.25s A = = = 42. 98 cm But y = Asin Therefore y = 42.98 sin = 0.02m (d) Direction of laser beam in the glass n = na = 1, ng = 1.5 ii) nasini = ngsinr therefore Sin r = 0.577 therefore r = 35.26o Cos 35.26 =, hyp = = 1.22cm But distance of displacement is given by sin 24.74 = Where d = 1.22sin24.74 = 0.51cm. (e) The principle of superposition state that when two or more waves move over each other, the outcome disturbance at some point through the medium usually could be determined from displacement produced by each wave this is a case of all linear systems. (f) Velocity for both pulses = 5ms-1 Distance at different times is given by velocity *time Therefore at 2sec 5*2 = 10m, at 3sec 5*3 = 15m, and at 4sec 5*4 = 20m The snapshots are as follows (g) The given harmonic frequencies are1875, 2625, and 3375 Hz. The difference between the given frequencies = 750. This implies that the pipe producing these frequencies is a pipe closed at one end. Pipes closed at one end only produces odd harmonics while a pipe that is open on both sides produces all harmonic. In this case, the successive resonant frequencies are odd therefore it is a closed at one end. (i) Car’s speed is 25ms-1, horn’s frequency is 300Hz, and speed of sound in air is 330ms-1 Wave equation V=fλ therefore, the wavelength of the horn = 330/300 = 1.1m Since the car is moving at a speed lower than that of sound, initially the driver will hear the horn sounding but as he moves closer, the sound will be reflected back by the warehouse and he will start hearing echoes of reverberation. The intensity of reverberation will increase as the car moves closer initially still clear but eventually the echoes gets mixed up as the sound hits the wall harder and being reflected at this point, the horn’s sound and echoes will overlap each other making the driver hear some deafening noise. At the point of impact, the driver will not hear the sound of impact but more overlapping of the sound creating a lot of noise. Section B 3. (a) Show that the linear wave equation is satisfied by the function y(x, t) = A sin(kx-ωt) Harmonic vibration at some point (x = 0) is given by y(t) = y(x= 0, t) = Asin(ωt) …...1 Where ω is angular frequency, t is a variable of time and A is the amplitude of vibration. This can be represented as in the figure below where for a specific time (t = 0), the associated harmonic wave function is y(x, t =0) = A sin …………2, lambda is the wavelength. Because the shape of the function should stay equal that is x – vt. Thus the x –t is based on wave function of which Eq. 2 changes to y(x, t) = A sin…...3 The period T of vibration is same as one in the wave. Recall tat T = 2п/ω = 1/v = λ/v. and v is wave frequency. Bringing wave constant k = 2п/ λ, the equation become Y(x, t) = Asin(kx- ωt) …...4 Due to wave motion, angular frequency can be given by ω = da/dt, where a is the angle from initial mark when t relating to some point in moving object. Integrating previous equation, it gives 2п = ωT, thus the expression of period T = 2п/ ω. Substituting with lambda λ= vT, and v is speed of given pulse. Thus v = λ/T = λ ω/2п. Hence we state new wave number k = 2п/ λ, and kv = ω. With x=vt, implying that ωt = ωx/vxT = 2пx/ λ = kx. And eventually, the wave function turns to y = f(x, t) = Asin t = Asint (b) Apply Newton’s second law to an element of a stretched string to show that the transverse velocity of waves on the string is given by v = (F/µ)1/2 When a string under tension is pulled to the side and released, the tension will act on acceleration of the given point backwards to equilibrium mark. Newton’s second law states tat acceleration of given point increases with raising tension. Increase in acceleration is due to increase in tension. Subsequently, wave speed reduces when mass per unit length of string is increased. Assuming the tension is F and mass per unit length is μ, then we can proof that v =. Dimension of F is ML/F2, and dimensions of μ are M/L. hence, the measurement F/μ are L2/F2 thus, measurement of are L/F, which is actually measurement of speed. Mechanically, pulse moving outwards with same speed v relative to fixed mark of reference is suitably to select a different point of reference. This alteration in reference point is accepted since Newton’s laws are legal for stationary or moving with constant speed. Now taking a small string segment ∆s and enlarged estimate arc of radius R. when changing reference point moving towards the right at speed v, the formed path has acceleration equivalent to v2/R that is distributed by elements of tension F in the string. The force F acts on both sides of partition and tangential to the arc. Horizontal elements of F cancel and all vertical elements F sinθ moves to the center creating radial force of 2F sinθ. Since the part is small, θ is also small, thus we apply small estimation θ≈θ thus, combined radial force is , the part has mass m = and since the part makes a circle subtends an angle 2θ at the middle, ∆s = R(2θ) and, therefore, m = μ∆s = 2μRθ From Newton’s second law using on this part, the radial element of movement achieves , 2Fθ = This equation is depended on preposition that pulse height is relatively small to length of string. From this preposition, we can estimate sin θ ≈ θ. In addition, since system assumes tension F is not influenced by existence of pulse, therefore, F is equal throughout the string. Hence, we can summarize that any shape travels on string has a velocity given by the equation v = . (c) Wave amplitude = 5 cm, period 0.3s Frequency will be given by f = 1/T = 1/0.3 = 3.333Hz Angular wave number k =2п/ λ, ω = 2п/ T = 2п/0.3 = 20.94 Maximum transverse velocity is from the wave form y(x, t) = Asin (kx±ωt) y(x, t) = 5sin(16+20.94) = 3ms-1 4) (a) sketch of a wave travelling on light strings and encounters a heavy boundary i.e. from Z1 to Z2 while the other case is vice versa of this. Z1 Z2 (b) Boundary conditions applying to two stretched strings When the strings posses the equivalent impedance (Z2=Z1) then, ρ = When the string is heavy for example a wall, Z2= and. Ρ= The outcome would be same amplitude waves leading to opposite directions that are characteristics for standing waves. When ρ=-1, the reflected wave zeros the incident wave at junction resulting to node at junction of the strings. When the 2nd string is very light then Z2/Z1 G ρ = This condition leads to standing wave however, because reflected wave sums to incident wave at junction there would be antinode at junction. When Z1> Z2 (heavy to light string) reflection coefficient is positive and reflected wave entails equal orientation like incident wave. When Z1< Z2 (light to heavy string) reflection coefficient is negative and reflected wave is spanned equivalent to incident wave. (c) Two strings at border condition, the transverse coordinate y and the derivative of transverse coordinate dy/dz should together be progressing. Taking the situation of incident on left border. Symbolize this wave by y1(z-vt). The wave shall be partly transmitted and partly reflected at the border. Assume the reflected wave be yR(z+vt) and the transmitted wave be yT(z-vt)…...1 Recalling z±vt shows the direction of spreading, then total wave in first string is, y1 = yT(z-vt) + yR(z+vt)……………………2 And total wave in second string is y2 = yT(z-vt)……………………3 Boundary conditions thus becomes yT(z=0, t) = + yR(z=0, t) = yT(z = 0, t)………4 …...5 These relations can be recast according to previous Eq. to become …………………6 The added different signs show the reflected waveform because of different direction of propagation. Further, the velocities are different. From Equ 5 -…………………7 Integrating this equation gives - yT(z=0,t)-yR(z=0,t)=…...8 However, integration should always be zero because when yT=0, yR and yT should, as well be zero. Thus, the tension in the two strings should be same when strings are not fixed between the strings Then Eq. 8 becomes yT(z=0,t)-yR(z=0,t)=…………9 Then using Eq. 5 and 9 yT(z=0,t)-yR(z=0,t)= …………………10 Therefore, we have resolved the reflection coefficient ρ that provides the fraction of incident wave front, which is reflected at border at some time. Equivalently, the transmission coefficient т could be derived from Equ. 5 and 10, yT(z=0,t) = yR(z=0, t) = yT(z=0, t) yT(z=0, t) + ρyT(z=0, t) = yT(z=0, t) = т ………………………….11 Read More