Question

I REALLY JUST WANT TO UNDERSTAND THIS TYPE OF PROBLEM - ANY HELP AT ALL ON THIS WOULD BE AMAZING:

Standing waves on a string are generated by oscillations having amplitude 0.005 m, angular frequency 942 rad/s, and wave number 0.750π rad/m.
a.) What is the equation of the standing wave?
b.) At what distances from x=0 are the nodes and antinodes?
c.) What is the frequency of a point on the string at an antinode?
d.) If the string is 4m long, how many nodes are there?

Answer 1

The equation of a wave is given as \[u(x,t) = A \sin(kx - \omega t + \psi)\]where u the amplitude, A is the maximum amplitude, k is the wavenumber, w is the angular frequency (rad/s), and psi is the phase shift. Can you come up with an expression of this wave using the above formula and the equation I have given you?

Answer 2

I would assume from your problem statement that 0.005m is the maximum amplitude. the value of u is the amplitude as we move along x. Obviously, this changes from 0.005m to -0.005m. If you were given amplitude at some x, wave number, angular frequency, and phase shift, we could calculate maximum amplitude by solving the wave equation for A.

psi and phi are the same here. It's just different notation. However, understand that both represent phase shift.

Are you doing calculus based physics or high school physics? I need to know how to approach part b.

Answer 3

Also, you're statement for the equation of the wave is absolutely correct. Great job! Remember, to keep the x and t variables. \[u(x,t) = 0.005 \sin(0.750 \pi x - 942t + {\pi \over 4} )\]

Answer 4

I am doing calculus based.

I have written down for node: x = nλ/2

and for antinode x= nλ/4

I'm just not sure where to go from there.

Also, THANK YOU so much for helping me out with this.

Answer 5

Those equations are valid. We could also find the locations of the anti-node by solving the derivative of the wave function for zero.

Let's use the handy equations here though. Let's define k, the wavenumber, in terms of wavelength, lambda. \[k = {2\pi \over \lambda}\]We can solve this equation for lambda. However, keep in mind that we have a phase shift, so actually, the location of the nodes are\[x_{node} = {\lambda \over 2} - \phi ~~ \text{and}~~ x_{anti-node} = {\lambda \over 4} - \phi\]Remember from algebra that when we add a number inside the sine function it shifts the graph left on the x-axis (in the negative direction). That is why we subtract the phase shift from the location of the nodes.

Answer 6

What do you get for the location of the node and anti-node? It may be helpful to set t=0 here.

Answer 7

We don't use frequency here. We need to use wavenumber. From the equation I presented earlier\[\lambda = {2\pi \over k} = {2\pi \over .750} = {8\pi \over 3}\]Therefore, \[x_{node} = {8\pi \over 6} - {\pi \over 4}\]

Answer 8

Oops. \[\lambda = {2\pi \over 0.75 \pi} = {8 \over 3}\]

Answer 9

ok, so by doing (8/3)/2 -π/4 I got .5479 for the x node

and (8/3)/4-π/4 I got -.1187 for the x antinode

Answer 10

Correct. One small correction. We can't have an anti-node behind the start of the wave. Let's do (16/12) - (pi/4) by setting n to be 2.

Answer 11

why did you choose to set n to 2?

Answer 12

Because it's the smallest value of n that makes the x value of the anti-node positive.

Answer 13

ok, that makes the anti-node .5479

Answer 14

Hmmm. I looked into standing waves. Are you sure we have a phase shift?

Answer 15

ummm no I am not sure. Also I am looking through my notes and I just came across the equation 2Asin(kx)cos(wt) written down. and there doesn't seem to be a phase shift...

Answer 16

Whoops! Let's regress here a bit. I was wrong in my analysis, but I've whipped out the old text book and gotten a handle on it. Sorry about that. x.x

For a standing wave, the initial wave travels through, then bounces off the opposite end. Therefore, we have two waves present in the string. These waves travel in opposite directions.

So we end up with \[y = 2Acos(\omega t)\sin(k x) = 0.01\cos(942 t)\sin(0.750 \pi x)\].

Now let's find the wavelength to be \[\lambda = {2 \pi \over k} = {2 \pi \over 0.75 \pi} = {8 \over 3}\]

Now we can get \[x_{node} = {8 \over 6} = {4 \over 3} ~~\text{and}~~ x_{anti-node} = {8 \over 12} = {2 \over 3}\]

Answer 17

http://en.wikipedia.org/wiki/Node_(physics) Take a look at the picture. Because it is a standing wave, we have a node at x=0 and x=4/3. The anti-node will have half way between these.

Answer 18

The problem statement says "nodes and anti-nodes." If we wanted to account for all of them, \[x_{node} = \left( 4 \over 3 \right) n ~~\text{and} ~~x_{anti-node} = \left(2 \over 3 \right) n\]

Answer 19

ok great. I am following this. Thanks for putting so much effort in to helping me.

Answer 20

No problem. Sorry that I got it wrong the first time around.

Answer 21

For part c. The frequency at the anti-node is related to the frequency of the wave. \[f = {\omega \over 2\pi}\]

Answer 22

I got f=942/2π = 149.9

Answer 23

Yep. The units on that is Hz, which is cycles/sec.

Answer 24

and for D would I just take 4 meters and divide that by (4/3) to get 3 nodes?

Answer 25

Plus the additional node at x = 0. Four nodes total.

Answer 26

Seriously, Thank you so much for helping me to understand this. I really appreciate it.

Answer 27

No problem. Sorry again about the confusion. It's been four years since I took this physics class.

Answer 28

No worries - and that's impressive that you were able to do that 4 years after the class.

For whatever reason I really can't grasp physics. Any of the other sciences - Chem, bio, A&P, I do really well - but I can't wrap my head around physics

Answer 29

I'm an engineering student. I use my physics everyday, just not the particulars of waves. It usually doesn't take me too long to re-learn it.

That happens. You seemed to grasp it really well. Keep practicing and try to think of things in your own terms. It will help you apply ideas from one problem to others.

Answer 30

Thanks again!

Answer 31

You're welcome! Let me know if you have any other questions.