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How to find the velocity of a wave in simple harmonic motion given time

Problem Statement
Find the speed of this longitudinal wave at t=0.12s.
Relevant Equations
v=w*xmax*cos(wt)
The graph provided is below. The problem asks for the speed of the wave at 0.12s. I used the formula v=w*xmax*cos(wt), provided in our textbook where xmax is the amplitude of 2 cm, w (omega) is 2pi divided by the period of 0.2. However, for some reason this formula doesn't give me the correct result. Instead, the solution to the problem involves the formula w*sqrt(xmax^2-x^2), which requires us to first find the displacement at t = 0.12.

My question is: why does the formula I used not work? Why must we use the other formula to solve this problem? I apologise for the bad formatting of the formulas.
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kuruman

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It doesn't work because it's the wrong formula to use. It predicts that the velocity is maximum at t = 0. If you look at the graph, the position is maximum at t = 0 which means that the speed is zero at t = 0.

Also, is this a wave or a simple harmonic oscillator? I think the latter.
 
It doesn't work because it's the wrong formula. It predicts that the velocity is maximum at t = 0. If you look at the graph, the position is maximum at t = 0 which means that the speed is zero at t = 0.

Also, is this a wave or a simple harmonic oscillator? I think the latter.
Why does the formula I used predict that velocity is maximum at t=0?
If I use the same formula but replacing cos with sin, I get the correct answer. Why is that? Is it related to what you said? We were taught to use cos when displacement is maximum at t = 0, and sine when displacement is 0 at t = 0. Is this correct?
 

kuruman

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You wrote
##v(t)=\omega~ x_{max}\cos(\omega~t)##
At ##t = 0##, ##v(0)=\omega~ x_{max}\cos(0)=\omega~ x_{max}(1)\neq 0.##
Try this
1. Find an expression for ##x(t)## consistent with the graph. Hint: As indicated above, ##x_{max}\cos(0)=x_{max}.##
2. Take the derivative with respect to time to find ##v(t)##. The result should answer your second question.
3. Evaluate ##v(t=0.12~s).## This method is neater and there is no need to do what the solution suggests.

We were taught to use cos when displacement is maximum at t = 0, and sine when displacement is 0 at t = 0. Is this correct?
That is correct. Your mistake was that you used cos for the velocity; you were taught to use cos for the displacement.
 

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