I am not sure how to work either of the following problems -- or where to begin for that matter.
1. A string of length 9 meters is fixed at both ends. It oscillates in its third harmonic with a frequency of 176 Hz and amplitude .31cm. What is the equation of motion of the point on the string which lies at 1.8 meters from the left end? What is the max. velocity of this point?
I worked a similar problem to this, but I am not sure how to deal with the third harmonic part of this.
See also the introductory problem sets and the text, in both of which I believe the equation y = A sin(kx) sin(omega t) is derived.
The third harmonic contains 3 half-waves in the length of the string and therefore has wavelength 6 meters.
When every point is at maximum displacement the shape of the wave is therefore given by the graph of
y = .31 cm sin( k x), where k is chosen so that the sine function goes thru a complete cycle when x changes by 6 m. Thus k * 6 m = 2 pi and k = 2 pi / (6 m) = pi / 3 m^-1, which is the wave number for this wave.
The frequency is 176 Hz, so every point undergoes SHM with amplitude .31 cm sin(kx) and angular frequency 2 pi rad / cycle * 176 cycles / sec = 352 pi rad /s. So the equation of the wave is
y(x, t) = .31 cm sin(pi/3 x) * sin(352 pi rad/s * t).
The radius of the reference circle for the SHM at position x = 1.8 m is .31 cm sin(pi/3 * 1.8 m) and the speed of the point on the reference circle is
ref circle speed = radius * angular frequency = amplitude at point * angular frequency = .31 cm sin(pi/3 * 1.8 m) * 352 pi rad / s. Very roughly, I think this is around 1000 cm / sec.
The velocity of the point on the wave matches the reference-circle speed when velocity is a maximum.
#2 Sound is created by the vibration of the air column in a pipe which is closed at one end and open at the other. The pipe is 3 meters long, and the speed of sound is 335 m/s. How long should a second pipe, also closed at one end and open at the other, be in order that its second harmonic match the frequency of the fundamental harmonnic of the first? We talked about these pipes in class but I am not clear on how to work this problem.
The first harmonic has just node and antinode, so 1/4 of the wavelength is equal to the length of the pipe and the wavelength is therefore 12 m.
The second harmonic has node, antinide, node, antinode, so 3/4 of the wavelength is equal to the length of the pipe. If L is the length of the second pipe, then the wavelength of the second harmonic is therefore 4/3 L.
To match frequencies, the wavelengths must be equal so we would have
4/3 L = 12 m and
L = 9 m.