course Phy 122 jhޫNassignment #012
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15:57:01 **** Query introductory set six, problems 11-13 **** given the length of a string how do we determine the wavelengths of the first few harmonics?
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RESPONSE --> When lambda decreases, 1/2 lambdas are formed. Using the harmonics 1, 2, 3 and multiply each by 1/2 lambda to determine the lambda for each harmonic. Ex: 1*1/2 lambda = L lambda = 2L
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15:57:56 ** As wavelength decreases you can fit more half-waves onto the string. You can fit one half-wave, or 2 half-waves, or 3, etc.. So you get 1 half-wavelength = string length, or wavelength = 2 * string length; using `lambda to stand for wavelength and L for string length this would be 1 * 1/2 `lambda = L so `lambda = 2 L. For 2 wavelengths fit into the string you get 2 * 1/2 `lambda = L so `lambda = L. For 3 wavelengths you get 3 * 1/2 `lambda = L so `lambda = 2/3 L; etc. } Your wavelengths are therefore 2L, L, 2/3 L, 1/2 L, etc.. FOR A STRING FREE AT ONE END: The wavelengths of the first few harmonics are found by the node - antinode distance between the ends. The first node corresponds to 1/4 wavelength. The second harmonic is from node to antinode to node to antinode, or 4/3. the third and fourth harmonics would therefore be 5/4 and 7/4 respectively. **
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RESPONSE --> lambda's are wavelengths ok
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16:00:00 **** Given the wavelengths of the first few harmonics and the velocity of a wave disturbance in the string, how do we determine the frequencies of the first few harmonics?
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RESPONSE --> Frequency is how many crests of waves pass per a certain amount of time. Frequency = velocity of wave / wavelength
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16:00:29 ** The frequency is the number of crests passing per unit of time. We can imagine a 1-second chunk of the wave divided into segments each equal to the wavelength. The number of peaks is equal to the length of the entire chunk divided by the length of a 1-wavelength segment. This is the number of peaks passing per second. So frequency is equal to the wave velocity divided by the wavelength. **
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RESPONSE --> okay
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16:03:57 **** Given the tension and mass density of a string how do we determine the velocity of the wave in the string?
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RESPONSE --> v = `sqrt (tension / (mass density)) v = `sqrt ( tension / (mass/length))
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16:04:04 ** We divide tension by mass per unit length: v = sqrt ( tension / (mass/length) ). **
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RESPONSE --> okay
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16:04:07 **** gen phy explain in your own words the meaning of the principal of superposition
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16:04:09 ** the principle of superposition tells us that when two different waveforms meet, or are present in a medium, the displacements of the two waveforms are added at each point to create the waveform that will be seen. **
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16:04:11 **** gen phy what does it mean to say that the angle of reflection is equal to the angle of incidence?
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16:04:13 ** angle of incidence with a surface is the angle with the perpendicular to that surface; when a ray comes in at a given angle of incidence it reflects at an equal angle on the other side of that perpendicular **
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16:04:15 query univ phy problem 15.48 (19.32 10th edition) y(x,t) = .75 cm sin[ `pi ( 250 s^-1 t + .4 cm^-1 x) ] What are the amplitude, period, frequency, wavelength and speed of propagation?
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16:04:17 ** y(x, t) = A sin( omega * t + k * x), where amplitude is A, frequency is omega / (2 pi), wavelength is 2 pi / x and velocity of propagation is frequency * wavelength. Period is the reciprocal of frequency. For A = .75 cm, omega = 250 pi s^-1, k = .4 pi cm^-1 we have A=.750 cm frequency is f = 250 pi s^-1 / (2 pi) = 125 s^-1 = 125 Hz. period is T = 1/f = 1 / (125 s^-1) = .008 s wavelength is lambda = (2 pi / (.4 * pi cm^-1)) = 5 cm speed of propagation is v = frequency * wavelength = 125 Hz * 5 cm = 625 cm/s. Note that v = freq * wavelength = omega / (2 pi) * ( 2 pi ) / k = omega / k. **
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16:04:20 **** Describe your sketch for t = 0 and state how the shapes differ at t = .0005 and t = .0010.
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16:04:22 ** Basic precalculus: For any function f(x) the graph of f(x-h) is translated `dx = h units in the x direction from the graph of y = f(x). The graph of y = sin(k * x - omega * t) = sin(k * ( x - omega / k * t) ) is translated thru displacement `dx = omega / k * t relative to the graph of sin(k x). At t=0, omega * t is zero and we have the original graph of y = .75 cm * sin( k x). The graph of y vs. x forms a sine curve with period 2 pi / k, in this case 2 pi / (pi * .4 cm^-1) = 5 cm which is the wavelength. A complete cycle occurs between x = 0 and x = 5 cm, with zeros at x = 0 cm, 2.5 cm and 5 cm, peak at x = 1.25 cm and 'valley' at x = 3.75 cm. At t=.0005, we are graphing y = .75 cm * sin( k x + .0005 omega), shifted -.0005 * omega / k = -.313 cm in the x direction. The sine wave of the t=0 function y = .75 cm * sin(kx) is shifted -.313 cm, or .313 cm left so now the zeros are at -.313 cm and every 2.5 cm to the right of that, with the peak shifted by -.313 cm to x = .937 cm. At t=.0010, we are graphing y = .75 cm * sin( k x + .0010 omega), shifted -.0010 * omega / k = -.625 cm in the x direction. The sine wave of the t = 0 function y = .75 cm * sin(kx) is shifted -.625 cm, or .625 cm left so now the zeros are at -.625 cm and every 2.5 cm to the right of that, with the peak shifted by -.625 cm to x = +.625 cm. The sequence of graphs clearly shows the motion of the wave to the left at 625 cm / s. **
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16:04:25 **** If mass / unit length is .500 kg / m what is the tension?
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16:04:27 ** Velocity of propagation is v = sqrt(T/ (m/L) ). Solving for T: v^2 = T/ (m/L) v^2*m/L = T T = (6.25 m/s)^2 * 0.5 kg/m so T = 19.5 kg m/s^2 = 19.5 N approx. **
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16:04:29 **** What is the average power?
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16:04:31 ** The text gives the equation Pav = 1/2 sqrt( m / L * F) * omega^2 * A^2 for the average power transferred by a traveling wave. Substituting m/L, tension F, angular frequency omeage and amplitude A into this equation we obtain Pav = 1/2 sqrt ( .500 kg/m * 195 N) * (250 pi s^-1)^2 * (.0075 m)^2 = .5 sqrt(98 kg^2 m / (s^2 m) ) * 62500 pi^2 s^-2 * .000054 m^2 = .5 * 9.9 kg/s * 6.25 * 10^4 pi^2 s^-2 * 5.4 * 10^-5 m^2 = 17 kg m^2 s^-3 = 17 watts, approx.. The arithmetic here was done mentally so double-check it. The procedure itself is correct. **
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t}ԑޯ assignment #013 013. `Query 11 Physics II 11-12-2008
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16:22:46 Query introductory set six, problems 15-18 how do we determine the energy of a standing wave given the amplitude and frequency of the wave and the mass of the string?
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RESPONSE --> Mass 'm' with amplitude 'A' and frequency 'f' has angular frequency `omega = 2 `pi f. Max velocity is `omega A = 2 `pi*f*A. KE = .5 m v^2 KE = .5 * (2 `pi * f * A)^2 KE = 2 `pi^2 * f^2 * A^2 * m KE = total energy total energy = 2 `pi^2 * f^2 * A^2 * m total mass of string = (mass/unit of length)*(string length) total energy of wave motion = 2 `pi^2 f^2 A^2 * `mu * L
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16:22:59 STUDENT ANSWER AND INSTRUCTOR RESPONSE: Energy = 2*pi^2*m*f^2*A^2 INSTRUCTOR RESPONSE: ** You should understand the way we obtain this formula. We assume that every point of the string in in SHM with amplitude A and frequency f. Since the total energy in SHM is the same as the maximum potential or the max kinetic energy, all we need to do is calculate the max potential energy or kinetic energy of each point on the string and add up the results. Since we know mass, frequency and amplitude, we see that we can calulate the max kinetic energy we can get the result we desire. Going back to the circular model, we see that frequency f and amplitude A imply reference point speed = circumference / period = circumference * frequency = 2 `pi A f. The oscillator at its maximum speed will match the speed of the reference point, so the maximum KE is .5 m v^2 = .5 m (2 `pi A f)^2 = 2 `pi^2 m f^2 A^2. **
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RESPONSE --> okay
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17:26:14 If the ends of two strings are driven in phase by a single simple harmonic oscillator, and if the wave velocities in the strings are identical, but the length of one string exceeds that of the other by a known amount, then how do we determine whether a given frequency will cause the 'far ends' of the strings to oscillate in phase?
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RESPONSE --> The mass density 'mu' of a string at tension 'T' gives that velocity 'v' = `sqrt (T/mu). The wave starts and ends at the same point along the string even though one string is longer than the other. The wave traveling on the longer string is behind the wave on the shorter string by the difference between those two strings, 'dL'. Constructive interference occurs when the wavelengths difference is a whole number then the waves reinforce. Destructive interference occurs when the difference of wavelengths is a whole number plus 1/2 a wavelength the waves cancel each other out. `lambda = v / f `lambda = 1/f * `sqrt(T / `mu). To determine if the strings oscillate in phase; # wavelengths = `dL / `lambda
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17:26:22 ** the question here is whether the far ends of the strings are at the same phase of motion, which occurs only if their lengths differ by exactly one, two, three, ... wavelengths. So we need to find the wavelength corresponding to the given frequency, which need not be a harmonic frequency. Any frequency will give us a wavelength; any wavelength can be divided into the difference in string lengths to determine whether the extra length is an integer number of wavelengths. Alternatively, the pulse in the longer string will be 'behind' the pulse in the shorter by the time required to travel the extra length. If we know the frequency we can determine whether this 'time difference' corresponds to a whole number of periods; if so the ends will oscillate in phase **
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RESPONSE --> okay
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17:26:25
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17:44:55 General College Physics and Principles of Physics 11.38: AM 550-1600 kHz, FM 88-108 mHz. What are the wavelength ranges?
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RESPONSE --> AM range: f = 550 kHz = 550 * 10^3 hz = 5.5 * 10^5 Hz `lambda = (3 * 10^8 m/s) / (5.5 * 10^5 cycles / s) `lambda = 545.5 m f = 1600 kHz = 1600 kHz * 10^3 hz = 1.6* 10^6 Hz `lambda = (3 * 10^8 m/s) / (1.6 * 10^6 cycles / s) `lambda =187.5 m FM range: f = 88.0 mHz = 88.0 * 10^6 Hz = 8.80 * 10^7 Hz `lambda = (3 * 10^8 m/s) / (8.80 * 10^7 cycles / s) `lambda = 3.41 m f = 108 mHz = 108 * 10^6 Hz = 1.08 * 10^8 Hz `lambda = (3 * 10^8 m/s) / (1.08 * 10^8 cycles / s) `lambda = 2.78 m
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17:44:57
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17:45:49 At 3 * 10^8 m/s: a frequency of 550 kHz = 550 * 10^3 Hz = 5.5 * 10^5 Hz will correspond to a wavelength of 3 * 10^8 m/s / (5.5 * 10^5 cycles / sec) = 545 meters. a frequency of 1600 kHz = 1.6* 10^6 Hz will correspond to a wavelength of 3 * 10^8 m/s / (1.6 * 10^6 cycles / sec) =187 meters. The wavelengths for the FM range are calculated similarly. a frequency of 88.0 mHz= 88.0 * 10^6 Hz = 8.80 * 10^7 Hz will correspond to a wavelength of 3 * 10^8 m/s / (8.80 * 10^7 cycles / sec) = 3.41 meters. The 108 mHz frequency is calculated similarly and corresponds to a wavelength of 2.78 meters.
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RESPONSE --> okay
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17:45:52
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17:47:36 General College Physics and Principles of Physics 11.52: What are the possible frequencies of a violin string whose fundamental mode vibrates at 440 Hz?
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RESPONSE --> 1st f = 440 Hz 2nd f = 2 * 440 Hz = 880 Hz 3rd f = 3 * 440 Hz = 1320 Hz 4th f = 4 * 440 Hz = 1760 Hz
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17:47:39
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17:47:47 The fundamental mode for a string fixed at both ends fits half a wavelength onto the string and therefore has a wavelength equal to double its length. The next three harmonics fit 2, 3 and 4 half-wavelengths into the length of the string and so have respectively 2, 3 and 4 times the frequency of the fundamental. So the first 4 harmonics are fundamental frequency = 440 Hz First overtone or second harmonic frequency = 2 * 440 Hz = 880 Hz Second overtone or third harmonic frequency = 3 * 440 Hz = 1320 Hz Third overtone or fourth harmonic frequency = 4 * 440 Hz = 1760 Hz
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RESPONSE --> ok
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17:47:50
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17:47:53 General College Physics Problem: Earthquake intensity is 2.0 * 10^6 J / (m^2 s) at 48 km from the source. What is the intensity at 1 km from the source?
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17:47:55
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17:47:57 The wave is assumed spherical so its surface area increases as the square of its distance and its intensity, which is power / surface area, decreases as the square of the distance. So the intensity at 1 km will be (48 km / 1 km)^2 = 2300 times as great, or 2300 * 2.0 * 10^6 J / (m^2 s) = 4.6 * 10^9 J/(m^2 s).
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17:47:59
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17:48:01 At what rate did energy pass through a 5.0 m^2 area at the 1 km distance?
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17:48:03
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17:48:06 Through a 5 m^2 area the rate of energy passage is therefore 4.6 * 10^9 J / (m^2 s) * 5.0 m^2 = 2.3 * 10^10 J / s, or 23 billion watts.
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