[SOLVED] The maximum and minimum transverse speeds of a point at an antinode


by TFM
Tags: antinode, maximum, minimum, point, solved, speeds, transverse
TFM
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#1
Feb28-08, 09:23 AM
P: 1,031
1. The problem statement, all variables and given/known data

Adjacent antinodes of a standing wave on a string are 15.0cm apart. A particle at an antinode oscillates in simple harmonic motion with amplitude 0.850cm and period 0.0750s. The string lies along the +x-axis and is fixed at x = 0.

the speed of the two travelling waves are 4.00ms

Find the maximum and minimum transverse speeds of a point at an antinode.

2. Relevant equations

I'm not sure

3. The attempt at a solution

Any help, I would have thought that they would have the same speed as the wave, ie 4 ms

TFM
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kkrizka
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#2
Feb28-08, 10:50 AM
P: 85
Simple harmonic motion means that the particle's vertical position can be given by the function y(t)=Asin(wt) where A is the max amplitude and w is the angular speed (2 pi / period). The deriative of that function will give you the speed as a function of time. Find the max and min of speed of that, and you are done. (Another hint: max of a absolute value of a cosine is 1 and min is 0).
TFM
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#3
Feb28-08, 12:48 PM
P: 1,031
[tex] v = A\omega cos (\omega t) [/tex]

I get

[tex] v = 0.0045 * (\frac{2 \pi}{0.075}) cos(\omega t) \omega[/tex]

I'm not sure what you meant by the cos part (I understand max absolute value = 1, minimuim = 0)

TFM

kkrizka
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#4
Feb28-08, 01:06 PM
P: 85

[SOLVED] The maximum and minimum transverse speeds of a point at an antinode


That seems about right. The cosine just means that the velocity will not be constant over time and will follow a cosine curve. However the question only asks for the maximum and minimum speeds, so all you have to do is find max and min of the function you found.

Btw, I think you have an extra w at the end.
TFM
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#5
Feb28-08, 01:22 PM
P: 1,031
Yeah, that extra w shouldn't be there.

I feel silly, but...

how do I work out the maximum/minimum of the function - don't you have to differentiate it again and find when it is equal to 0?

TFM (Rather embarrased )
kkrizka
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#6
Feb28-08, 01:25 PM
P: 85
That's where the maximum and minimum of the cosine comes in. Since the velocity only changes as a function of time. And as time changes the only thing that changes is the cosine. So the function will be at a maximum when cosine as at a maximum and at minimum when consine is at minimum.
TFM
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#7
Feb28-08, 01:50 PM
P: 1,031
Lets see if I understand, the maximum will be when:

[tex] v = 0.0045 * (\frac{2 \pi}{0.075}) cos(1) [/tex]

And a minimum when:

[tex] v = 0.0045 * (\frac{2 \pi}{0.075}) cos(0) [/tex]

?

TFM
matthewpowers
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#8
Feb28-08, 02:28 PM
P: 16
Quote Quote by TFM View Post
Lets see if I understand, the maximum will be when:

[tex] v = 0.0045 * (\frac{2 \pi}{0.075}) cos(1) [/tex]

And a minimum when:

[tex] v = 0.0045 * (\frac{2 \pi}{0.075}) cos(0) [/tex]

?

TFM
Not quite. The velocity will be largest when [cos(wt)] is at it's largest (the whole cosine function, not whats inside the cosine function, since what's inside the function can get very large if time gets very large. But no mater how big (wt) gets, the cosine function never gets larger than a certain value, and it is the value of the whole cosine function that multiplies the other terms in the equation to give the velocity), and the velocity will be at a minimum when cos(wt) is at it's smallest (absolute value). What is the largest value that a cosine function ever reaches? What is the smallest (absolute) value that cosine ever reaces?
TFM
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#9
Feb28-08, 02:39 PM
P: 1,031
Ah, Should be:

[tex] cos (\omega t) = 0 [/tex] for minima

[tex] cos (\omega t) = 1 [/tex] for maxima

that gives 0 and pi/2, though? do you use 2 pi?

TFM
matthewpowers
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#10
Feb28-08, 02:50 PM
P: 16
right, wt=0 -> velocity is max, wt=pi/2 -> velocity is minimum. What are those maximum and minimum velocities?
TFM
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#11
Feb28-08, 02:53 PM
P: 1,031
I used pi/2 and 2pi as the values, which gave me 2.37 and 0.59, but they were stated as being wrong

TFM
matthewpowers
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#12
Feb28-08, 03:17 PM
P: 16
And your calculator is in radian mode right. :)
matthewpowers
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#13
Feb28-08, 03:19 PM
P: 16
You should be able to use zero, just as well as 2pi...
TFM
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#14
Feb28-08, 03:27 PM
P: 1,031
I'm using excel, which operates inj radians

I did

cos(wt) = 0 for minima, giving pi/2
cos(wt) = 1 for maxima giving 0, 2pi

then

using:

v = Awcos(wt)

A = 0.00425 m (0.425cm)
w = 2pi/0.075 (period = 0.075 s)

[tex] v_m_i_n = 0.00425*(2pi/0.075)*(pi/2) [/tex] = 2.23
[tex] v_m_a_x = 0.00425*(2pi/0.075)*(2pi) [/tex] = 0.559

Any ideas?

TFM
matthewpowers
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#15
Feb28-08, 03:51 PM
P: 16
The pi/2 and 0,2pi are the values of wt that make the cosine function 0 and 1, respectively. By doing what you did, you are saying v = Aw*(wt), as opposed to v=Awcos(wt).
TFM
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#16
Feb28-08, 04:19 PM
P: 1,031
I did v = Awcos(pi/2), Awcos(0), which gave me a maximum of 0.356, and a minimum of 2.18 x 10^-17m, which is wrong!?!?

TFM
matthewpowers
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#17
Feb28-08, 04:42 PM
P: 16
Well, I think one of them is right...I'm pretty darn sure the min is zero (which is what 2.18x10^-17 is).

What are the correct answers?
matthewpowers
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#18
Feb28-08, 05:05 PM
P: 16
Oh, and your aplitude should be .0085, not .00425, so you would be getting half the correct answer for the max...


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