Phy 202
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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This is the question and answer I received previously:
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 .31 cm. 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 maximum velocity of this point?
For this question, I just dont know what it means to find the equation of motion. Would it be v=(amp)(freq)(2pi) and if so why does it matter whether the point is at 1.8m?
This is covered in the assigned Introductory Problem Sets.
When the wave is at maximum displacement its y vs. x graph is y = A sin(k x).
Thus the particle at point x undergoes simple harmonic motion with amplitude A_x = A sin(kx).
Simple harmonic motion with amplitude A_x is described by the y vs. t function y = A_x sin(omega t).
The complete equation of motion includes both the x and y dependencies, so we replace A_x with A sin(kx) to get
y = A sin(kx) sin(omega t).
Note that you haven't yet covered simple harmonic motion in Physics I. If you aren't familiar with simple harmonic motion, which is the last topic covered in that class, you will have trouble understanding what's going on here. You might want to have a good look at Introductory Problem Set 9.
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Ok so i understand this a bit better and I have looked at the problem set 9 stuff a bit ( i plan on looking a bit closer). I still have a small question that is mostly technical.
So once I figure out the equation I am supposed to use is
y = A sin(kx) sin(omega t), I am assuming I should solve for y? If so, I know that omega is frequency*2pi, A=.31cm converted to m, but what is k we dont know time. I guess I am asking what you would expect for a final answer to this problem. Would it just be the equation you displayed above or would I need to figure out f and t?
The rest of the help you sent was amazing! Thanks!
I posted this last night, by accident, without a response. I apologize for that.
y = A sin(kx) sin(omega t) is a function of position x and clock time t. To emphasize this let's use function notation and write
y(x, t) = A sin(kx) sin(omega t).
For the given problem the amplitude is A = .31 cm.
sin(kx) completes a cycle when kx changes by 2 pi. The third harmonic of that string has wavelength 6 meters, so k x has to change by 2 pi when x changes by 6 meters , so k * 6 meters = 2 pi radians and k = 2 pi rad / (6 meters) = pi / 3 rad / m.
sin(omega t) completes a cycle when omega t changes by 2 pi. The frequency is 176 Hz, so omega t has to change by 2 pi when t changes by 1/176 second. Thus omega * 1/176 sec = 2 pi rad and omega = 352 pi rad / sec.
The equation is therefore
y (x, t) = .31 cm sin(pi/3 rad/m * x) sin(352 pi rad/s * t).
The x = 1.8 meter point therefore has equation of motion
y(1.8 m, t) = .31 cm sin(pi/3 rad/m * 1.8 m) sin(352 pi rad/s * t).
sin(pi/3 * 1.8) = .95, approx. so .31 cm sin(pi/3 rad/m * 1.8 m) = .31 cm * .95 = .30 cm, approx. This is interpreted as the amplitude of the motion at the x = 1.8 m point. At that point the equation of the simple harmonic motion is
y(t) = .30 cm sin(352 pi rad/s * t).
The particle at the y = 1.8 m point undergoes simple harmonic motion with amplitude .3 m, much like that of a pendulum but with a much higher frequency (also much like a mass on a stiff spring, which might well have this frequency).
You might well have more questions, which are always welcome.