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Finding the maximum velocity of a wave on a tight string

  • Thread starter lol34543
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Homework Statement


If the displacement of a tight string is represented by

y(x,t)= Acos(2∏/λ(x-vt))

Determine an expression for the velocity vy at which a section of the string travels. What is the maximum value of Vy? When is this maximum value greater than the wave propagation speed v?

The Attempt at a Solution



I started by differentiating the equation to get Vy = -A(2∏/λ)vsin((2∏/λ)(x-vt))
I then said that Vy would reach a maximum when sin((2∏/λ)(x-vt)) = 1 but I don't think this is right. Any help would be appreciated. Thank you.
 

Answers and Replies

  • #2
Simon Bridge
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I then said that Vy would reach a maximum when sin((2∏/λ)(x-vt)) = 1 but I don't think this is right.
Why not? It gives a maximum speed of Aω
 
  • #3
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This is incorrect:

y(x,t)= Acos(2∏/λ(x-vt))

[itex] y(x,t)=Acos(\frac{2 ∏}{λ}(x-vt)) [/itex]

Then you have differentiated. With respect to t :

Vy = -A(2∏/λ)vsin((2∏/λ)(x-vt))

[itex] \frac{d y(x,t)}{dx}= -A(\frac{2∏}{λ}) v sin(\frac{2∏}{λ}(x-vt)) [/itex]

However, I think that you may have a small sign error. The -v should give you one minus sign, but you also differentiated cos(x) which gives -sin(x)

So:

[itex] \frac{d y(x,t)}{dx}= v_{y}= A(\frac{2∏}{λ}) v sin(\frac{2∏}{λ}(x-vt)) [/itex]

So far so good. Then by stating that the [itex] v_{y}=1 [/itex] you are effectively stating that you believe the maximum value that Vy can be is 1.

This may be correct in some contexts, but my advise would be to consider how you would find the maximum of a function using differentiation?
 
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  • #4
Simon Bridge
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So far so good. Then by stating that the vy=1 you are effectively stating that you believe the maximum value that Vy can be is 1.
I don't think OP said that, I thought he said:

"I then said that Vy would reach a maximum when sin((2∏/λ)(x-vt)) = 1 "


So: the max v would be A(2∏λ)v wouldn't it?
 
  • #5
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My apologies for my mistake you are indeed correct. Please ignore my ramblings.
 
  • #6
Simon Bridge
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No worries, done that myself.
Checking via differentiation was good advise though.
 

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