Finding location of nodes and antinodes.

[blueskadoo42]

Homework Statement

Two transverse waves traveling on a string combine to a standing wave. The displacements for the traveling waves are Y1(x,t) = 0.0200 m sin(2.00 m−1 x − 2.90 s−1 t + 0.40) and Y2(x,t) = 0.0200 m sin(2.00 m−1 x + 2.90 s−1 t + 0.80), respectively, where x is position along the string and t is time.
Find the location of the first antinode for the standing wave at x > 0.
Find the first t > 0 instance when the displacement for the standing wave vanishes everywhere.

Homework Equations

The Attempt at a Solution

I got the first part. The correct answer was .485 m. Now I wasnt sure if you should take the location where x is a maximum(anitnode) or where x is a minimum(node). Any help would be greatly appreciated. Thank you.

 

[Redbelly98]

At a node, the displacement Y is zero at all times, so that won't help with solving the problem. Try using the antinode location you found.

 

[blueskadoo42]

negative. the antinode I found does not work for when the standing wave vanishes completely. thanks for the thought.

 

[Redbelly98]

That's weird, I was able to find t that makes Y1+Y2=0 at the antinode x=0.485.

Oh well, good luck.

 

[blueskadoo42]

Yeah i found a t of 2.23s at that x. Finding that t value is to find where sin would go to 0 right? am i understanding that correctly? that is where the wave will vanish right?

 

[blueskadoo42]

bump. anyone?

 

[Redbelly98]

Yeah i found a t of 2.23s at that x. Finding that t value is to find where sin would go to 0 right? am i understanding that correctly? that is where the wave will vanish right?

You're thinking is correct, for x at an antinode. However, for your value of t we get:

sin(2*0.4854 - 2.9*2.23 + 0.4)
= sin(-5.096 radians)
= 0.927, not zero.

Maybe you made an algebra error somewhere?

 

[blueskadoo42]

that might be it. but let me make sure I comprehend what you have just stated. That .927 is the position where the wave vanishes? not the time? Or is that the time where the wave would vanish? This one has got me wound up hah

 

[Redbelly98]

0.927 is the value of the sine function, for x=0.4854m and t=2.23s.

We want to find the values of t that will make that sine = 0, and then choose the smallest positive t for which that is true.

 

[blueskadoo42]

alright. so the sin(2pi)=0 so set the equation equal to 2pi. solve for t, and that's the time? What about wavelength, how would I know if that's the smallest t? I tried graphing it on excel but I am not sure what to keep constant when solving for time.

 

[blueskadoo42]

also, do I use the combined equation?

 

[blueskadoo42]

err. sin(pi)=0. I am using that one

 

[blueskadoo42]

I solved for t=.611s. so when i plug in;

sin(2*.485 + 2.9*.611 + .4) => about equal to sin(3.1419) => about equal to sin(pi)=0

is there where the wave would vanish, pending no mathematical error/ is this the right equation to use?

 

[Redbelly98]

sin(2*.485 + 2.9*.611 + .4)

This is not one of the original expressions.

sin(2pi)=0 so set the equation equal to 2pi

There are also other numbers besides 2pi that give sin(number)=0

 

[blueskadoo42]

pi and 0? pi would be the one to choose. so set either equation? equal to pi and solve for t. that's is where I am getting tripped up at. which equation to use. does it make a difference because they have different phases?

 

[Redbelly98]

pi and 0? pi would be the one to choose.

Those, yes, and also infinitely many others including -pi, -2pi, etc.
Examine the values of t you get for each, before deciding which to choose.

... so set either equation? equal to pi and solve for t. that's is where I am getting tripped up at. which equation to use. does it make a difference because they have different phases?

It shouldn't matter which equation, I happened to use the first one and then checked that answer in the 2nd equation to be sure.

Also, did you catch the error I pointed out at the beginning of post #14?