course Phy 202
.................................................??????????€??
......!!!!!!!!...................................
00:04:51 **** Query introductory set six, problems 11-13 **** given the length of a string how do we determine the wavelengths of the first few harmonics?
......!!!!!!!!...................................
RESPONSE --> We use the formula x*.5*'lambda=L, where x is the number of wavelengths that will fit on a string and L is the length of the string. The equation can then be rearranged depending on what is being solved for. Also, x depends on whether one wavelength fits on the string, or 2, or 3. This number is what is filled in for x.
.................................................
......!!!!!!!!...................................
00:06:44 ** 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. **
......!!!!!!!!...................................
RESPONSE --> I understand the first part; however, the second part regarding a ""string free at one end"" is confusing. I understand the relationships between the nodes and antinodes in all of them except the 4/3 for the second. I Just don't understand where the three comes from.
.................................................
......!!!!!!!!...................................
00:07:49 **** 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?
......!!!!!!!!...................................
RESPONSE --> The frequency is how many crests pass a certain point during a certain time frame.
.................................................
......!!!!!!!!...................................
00:08:02 ** 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. **
......!!!!!!!!...................................
RESPONSE --> This makes sense.
.................................................
......!!!!!!!!...................................
00:09:06 **** Given the tension and mass density of a string how do we determine the velocity of the wave in the string?
......!!!!!!!!...................................
RESPONSE --> v = sqrt (tension / (mass/length))
.................................................
......!!!!!!!!...................................
00:09:26 ** We divide tension by mass per unit length: v = sqrt ( tension / (mass/length) ). **
......!!!!!!!!...................................
RESPONSE --> This is about as straightforward as it seems to get.
.................................................
......!!!!!!!!...................................
00:10:45 **** gen phy explain in your own words the meaning of the principal of superposition
......!!!!!!!!...................................
RESPONSE --> I think that when two waves meet they are added at the points where they meet?
.................................................
......!!!!!!!!...................................
00:11:03 ** 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. **
......!!!!!!!!...................................
RESPONSE --> Ah, the *displacements* are added. That makes a little more sense.
.................................................
......!!!!!!!!...................................
00:12:14 **** gen phy what does it mean to say that the angle of reflection is equal to the angle of incidence?
......!!!!!!!!...................................
RESPONSE --> I'm not sure.
.................................................
......!!!!!!!!...................................
00:13:16 ** 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 **
......!!!!!!!!...................................
RESPONSE --> Basically, when an angle comes in on one side, it ""comes through"" or ""shines through"" at the same angle on the other side.
.................................................
......!!!!!!!!...................................
00:13:19 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?
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
00:13:20 ** 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. **
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
00:13:23 **** Describe your sketch for t = 0 and state how the shapes differ at t = .0005 and t = .0010.
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
00:13:24 ** 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. **
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
00:13:26 **** If mass / unit length is .500 kg / m what is the tension?
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
00:13:28 ** 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. **
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
00:13:29 **** What is the average power?
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
00:13:31 ** 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. **
......!!!!!!!!...................................
RESPONSE -->
.................................................
"