course Phy 202 ???????D?€???assignment #021
......!!!!!!!!...................................
15:23:50 **** 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 --> The first harmonic is 1/2 *lambda = L so lambda = 2L second harmonic is 2/2 *lambda = L so lambda = L third harmonic is 3/2 * lambda = L so lambda = (2/3)L The pattern continues in the same manner as illustrated above.
.................................................
......!!!!!!!!...................................
15:26:21 ** 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 --> For wavelengths traveling on a string with one free end the there is 1/4 wavelength between the node and antinode so then the rest are found along the principle that there is 1/4 wavelength between each node and antinode so the second harmonic is 4/3 and etc.
.................................................
......!!!!!!!!...................................
15:27:45 **** 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 --> Frenquency = velocity of a wave disturbance/wavelength F= v/lambda
.................................................
......!!!!!!!!...................................
15:27:53 ** 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 --> Ok
.................................................
......!!!!!!!!...................................
15:28:27 **** Given the tension and mass density of a string how do we determine the velocity of the wave in the string?
......!!!!!!!!...................................
RESPONSE --> Use the equation: v = sqrt(tension/(mass per unit length))
.................................................
......!!!!!!!!...................................
15:31:09 **** gen phy explain in your own words the meaning of the principal of superposition
......!!!!!!!!...................................
RESPONSE --> Ok. the principal of superposition is when two waveforms meet and the resulting displacement is the sum of their individual displacements.
.................................................
......!!!!!!!!...................................
15:31:24 ** 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 --> ok
.................................................
......!!!!!!!!...................................
15:33:26 **** gen phy what does it mean to say that the angle of reflection is equal to the angle of incidence?
......!!!!!!!!...................................
RESPONSE --> The angle of incidence is the angle that is made between the incident ray from the wave front and the reflecting surface. It is equal then to the angle of reflection because it is the corresponding angle of the reflected wave.
.................................................
......!!!!!!!!...................................
15:33:34 ** 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 --> ok
.................................................
......!!!!!!!!...................................
15:33:42 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 --> not assigned
.................................................
......!!!!!!!!...................................
15:33:47 ** 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 --> ok
.................................................
"