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Wave- displacement for a point

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Homework Statement



A wave on a string has a wave function with the form of:
y(x,t)=0.02sin(6.35x+2.63t)

Find the shortest time for the displacement of the string at any point on the chain changes from 0.01m to -0.02m.

The Attempt at a Solution



The answer said we can set x=0.
And 0.01=0.02sin(2.63t1), 3*3.14/2=2.63t2, where t1 is initial time and t2 is final time.
then t1=0.2s and t2=1.6s, so t2-t1=1.4s

But I don't understand where 3*3.14/2=2.63t2 comes from.
And when I tried to follow the answer and calculate, the calculated answers don't seem to be correct....

So how should I solve it?
Thank you very much.
 

Answers and Replies

  • #2
ehild
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But I don't understand where 3*3.14/2=2.63t2 comes from.
And when I tried to follow the answer and calculate, the calculated answers don't seem to be correct....

So how should I solve it?
Thank you very much.
At t2, the displacement is -0.02 at x=0. 0.02 sin(2.63 t2)=-0.02--> sin(2.63 t2)=-1. What is the angle if its sine is -1?

ehild
 
  • #3
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At t2, the displacement is -0.02 at x=0. 0.02 sin(2.63 t2)=-0.02--> sin(2.63 t2)=-1. What is the angle if its sine is -1?

ehild
270 degree, and why does the angle matter please?
 
  • #4
BruceW
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well, you have the equation 0.02 sin(2.63 t2) = -0.02 And you want to find t2, right? So what is the usual way to find t2 from here?

Also, remember there are many possible values for the angle, so there are many possible values for t1 and t2. But they want you to find values for t1 and t2 such that t2-t1 is as close to zero as possible.
 
  • #5
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well, you have the equation 0.02 sin(2.63 t2) = -0.02 And you want to find t2, right? So what is the usual way to find t2 from here?

Also, remember there are many possible values for the angle, so there are many possible values for t1 and t2. But they want you to find values for t1 and t2 such that t2-t1 is as close to zero as possible.
I got that now!!!!
That means 2.63 t2 = 3/4∏, right?
But still, the calculated t2 is 1.79, which is not 1.6 from the solution. Anything wrong?
 
  • #6
BruceW
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There are many possible correct pairs of t1 & t2. In the solution, they use the pair 0.2s & 1.6s
You have found 1.79s as a solution for t2. So you need to find a value for t1 which matches this value of t2.
OR, you can just try to find t2=1.6s instead. So, your equation was sin(2.63 t2) = -1 Can you find some other angles that will work here?
 
  • #7
BruceW
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hey, wait a minute.. I think their pair 0.2 & 1.6 does not work. Yeah, I'm pretty sure the solution is wrong. Anyway, keep going with your solution. You have got a value for t2. So now you need to find a value for t1.
 

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