course PHY202
I just wanted you to look over these and make sure I did them right.Thanks.
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1. Traveling waves are set up in a pair of long string by a single harmonic oscillator. The strings have identical mass densities of 8.8 g/m. Both strings terminate at a short bungee cord attached to a wall. The harmonic oscillator is attached to the other ends of the strings in such a way that one string is 7.6 m longer than the other. If the oscillator has f= 71 Hz. Give at least two string tensions which will produce the maxium motion of the bungee cord, and give at least two string tensions which will produce the minimum motion of the bungee cord.
2. Find the frequencies of the first four harmonics of a standing sound wave in an open pipe of length 6 m and in which sound propagates at 340 m/s.
3. A string of length 9m is fixed at both ends. It oscillates in its fourth harmonic with a frequency of 140 Hz and amplitude .24 cm. What is the equation of motion of the point on the string which lies at 3.4m from the left end. What is the maximum velocity of this point?
4. What is the angle of total internal reflection from a material which has index of refraction 1.75 to a material with index of refraction 1.09?
1. Max: vprop= 7.6 * 71Hz= 539.6m/s,
Ft=v^2*m/L = (539.6m/s)^2 * .0088kg = 2562.3N and
vprop= 3.8 * 71Hz = 269.8m/s,
Ft = (269.8 m/s)^2 * .0088kg = 640.5 N
Min: vprop = 15.2 * 71Hz = 1079.2m/s,
Ft= (1079.2m/s)^2 * .0088kg = 10249N and
vprop = 5.07 * 71Hz = (359.9m/s)^2,
Ft= (359,9m/s)^2 * .0088kg = 1139.8N
2. lambda = v/f = 340 m/s / 28.3 Hz = 12.0 m , fundamental harmonic f = v/2L = 340 m/s / 2* 6m = 28.3 Hz ,
2nd harmonic 1/2 lambda = 1/2 12 = 6 f = v/lambda = 340m/s / 6m = 56.6 Hz ,
3rd harmonic 1 lambda = 12 f= 340m/s /12m = 28.3 Hz ,
4th harmonic 3/2 lambda = 3/2 12 = 18 f = 340 m/s / 18 m = 18.8 Hz
For the fundamental you have 1/2 lambda = length, so lambda = 2 * length.
For the nth harmonic you have n * 1/2 labmda = length, so lambda = 2 / n * length.
So for example, for the third harmonic we have 6 node-antinode distances, corresponding to 6 * 1/4 lambda or 3/2 lambda, equal to the length. Thus
3/2 lambda = length and
lambda = 2/3 * length = 2/3 * 12 m = 8 m.
3. I got a little confused on this one. Since its closed on both ends it would have a Node on the ends an ananodes in between.
Would I need to find the vprop first and then go from there?
Right. The first harmonic, or fundamental, has configuration N A N.
The second would be N A N A N, etc..
From these configurations you find the wavelengths.
You do need the propagation velocity to find the frequency of each wavelength.
4. theta critical = sin -1 1.09/1.75 = .672 , since it reflects all of the light that would be all I needed to do right?
Right.