
#1
Feb2808, 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 +xaxis 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 



#2
Feb2808, 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).




#3
Feb2808, 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 



#4
Feb2808, 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. 



#5
Feb2808, 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 ) 



#6
Feb2808, 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.




#7
Feb2808, 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 



#8
Feb2808, 02:28 PM

P: 16





#9
Feb2808, 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 



#10
Feb2808, 02:50 PM

P: 16

right, wt=0 > velocity is max, wt=pi/2 > velocity is minimum. What are those maximum and minimum velocities?




#11
Feb2808, 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 



#12
Feb2808, 03:17 PM

P: 16

And your calculator is in radian mode right. :)




#13
Feb2808, 03:19 PM

P: 16

You should be able to use zero, just as well as 2pi...




#14
Feb2808, 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 



#15
Feb2808, 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).




#16
Feb2808, 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 



#17
Feb2808, 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? 



#18
Feb2808, 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...



Register to reply 
Related Discussions  
Maximum and minimum value  Calculus & Beyond Homework  8  
[SOLVED] minimum and maximum values  Calculus & Beyond Homework  11  
[SOLVED] Maximum and minimum problem  Calculus & Beyond Homework  5  
Maximum speed of a transverse wave  Advanced Physics Homework  2  
Maximum/Minimum Problem  Calculus  9 