6
$\begingroup$

I'm studying for my upcoming AP Physics 1 exam but can't figure out this problem

enter image description here

A student shakes a horizontally-stretched cord, creating waves. The graph above shows the vertical position $y$ as a function of time $t$ for a point on the cord. The student then tightens the cord so that waves on it will travel faster than before. How should the student now shake the cord to make the graph of $y$ versus $t$ for the point look the same as above?

(A) With fewer shakes per second than before

(B) With the same number of shakes per second as before

(C) With more shakes per second than before

(D) The answer cannot be determined without knowing the wavelength of the waves.

My intuition would tell me that increasing the speed of the waves would cause the point to oscillate at a faster rate vertically, thus fewer shakes per second than before are needed to maintain the same frequency for the particle's oscillation. However, the correct answer is B, so I really need a thorough explanation as to why the answer is B.

Improve this question
$\endgroup$
1
  • 1
    $\begingroup$ If you shake it 3 times per second the whole thing shakes at 3 times per second, not 6, not 9, not 100. $\endgroup$
    – user253751
    Apr 28 at 10:42

4 Answers 4

Reset to default
7
$\begingroup$

It's not specified which point on the cord we are focusing on. That is for good reason: different points behave the same way, just with a different phase offset, which is not relevant to the question.

As long as we are picking an arbitrary point, why not pick the end point in the hand of the student? The speed with which the waves travel then clearly does not affect the graph.

Improve this answer
$\endgroup$
5
$\begingroup$

You already have a nice, brief and concise answer but I'd like to add more to it.

From the "fundamental" equation for waves

$$c = \lambda f \tag{1}\label{1}$$

where $c$ is the speed of wave propagation, $\lambda$ the wavelength and $f$ the temporal frequency, we can see that if you change the speed of propagation then the right-hand side must also change for the equality to hold. There are two quantities that could change in an arbitrary way to provide the result we need, so this is an under-determined problem.

What comes to rescue is the fact that the frequency of oscillation depends solely on the external excitation force (the student that shakes the chord in this case) (please keep in mind that this is the case for linear waves). Thus, even if the speed of propagation changes, this won't affect the frequency. So, what has to change is the wavelength, which will change in order to accommodate the necessary changes for the equality to hold.

To provide some more intuition, consider the fact that we refer to a specific point in space. Thus, wavelength, which encodes information of the spatial variations of the wave, is completely irrelevant to our problem.

Improve this answer
$\endgroup$
3
$\begingroup$

Since this is a $y$ vs. $t$ graph, the frequency of the wave can easily be picked out as the (inverse) time between matching parts on the wave (e.g. peak to peak, $0$ to $0$, trough to trough, etc.). In this case our analysis does not depend on the wave speed, which relates to the rate at which the wave travels through space. If we want the same $y$ vs. $t$ graph, then we need the same frequency.

Improve this answer
$\endgroup$
2
$\begingroup$

Bottom line intuitive answer: Changing the propagation speed affects the relationship between frequency and wavelength. The wave moves faster, but is correspondingly longer, and those two cancel out, leaving the frequency the same at any point on the cord.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.