course Phy 232 NTz[{Qxassignment #021
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20:04:35 **** 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 --> first harmonic is when half the wave fits on the string so it will take 2 times the length for one full wave second harmonic is when 2 halves the wave fit on the string, so it takes 1 length for one wave third harmonic is when 3 halves fit on the length which means it takes 2/3 the length for one wave copied my response from assign 19
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20:53:59 ** 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 --> I tried reading up on the free ended string, but I still don't understand...... wouldn't the second harmonic be node to antinode to node and therefore 1/2 wavelength??
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20:54:57 **** 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 --> freq = velocity/lambda - so we divide the wavelength of each harmonic into the velocity
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20:55:16 ** 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 --> copied from 19
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20:55:27 **** 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 --> equation 15.13 v = sqrt(F/mu) = sqrt(tension/mass density)
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20:55:33 ** We divide tension by mass per unit length: v = sqrt ( tension / (mass/length) ). **
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RESPONSE --> copied
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20:56:22 **** gen phy explain in your own words the meaning of the principal of superposition
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RESPONSE --> adding amplitudes of two waves to get resulted wave
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20:56:59 ** 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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RESPONSE --> copied
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20:57:20 **** gen phy what does it mean to say that the angle of reflection is equal to the angle of incidence?
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RESPONSE --> looking at the reflection will give you the same angle from the reflective surface as if you could see the incidence.
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20:57:31 ** 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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RESPONSE --> copied
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21:08:31 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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RESPONSE --> Amplitude is .75 cm freq is 250 pi/2 pi = 125 period is 1/125 = .008 wavelength is 2 pi / .4 pi = 5 speed is 125 * 5 = 625
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21:08:40 ** 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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RESPONSE --> copied
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21:32:57 **** If mass / unit length is .500 kg / m what is the tension?
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RESPONSE --> using velocity of 625 => 6.25 m and .5, we get: 6.25^2 * .5 = 19.5
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21:33:07 ** 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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RESPONSE --> copied
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21:33:43 **** What is the average power?
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RESPONSE --> eq 15.25 Pav = 1/2 sqrt(mu * tension * omega pi^2 * amplitude^2) 1/2 sqrt(.5 * 19.5 * 250^2 * .0075^2) = 5.85 watts
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21:34:02 ** 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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RESPONSE --> copied and added pi to omega ^2
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