Need help with a few wave problems

  • Thread starter feelau
  • Start date
  • #1
61
0
Hi, so here's the problem: A transverse wave on a string is described by the following equation. y(x, t) = (0.35 m) sin[(1.25 rad/m)x + (108.2 rad/s)t]
Consider the element of the string at x = 0. What is the time interval between the first two instants when this element has a position of y = 0.303 m? since the coefficient in front of the t variable is angular frequency which is equaled to 2pi/period, I thought I could just solve for the period and that is the time difference regardless of its position, but it's not correct so could anyone tell me how to go about solving this problem? I assume that my assumptions are flawed hehe.... thank you for helping
 

Answers and Replies

  • #2
OlderDan
Science Advisor
Homework Helper
3,021
2
Except for the points at which |y| is the amplitude, every y value is hit twice during each period of oscillation.
 
  • #3
61
0
so they actually mean half of the period?
 
  • #4
OlderDan
Science Advisor
Homework Helper
3,021
2
feelau said:
so they actually mean half of the period?
Not likely. Points that have the same displacement are not equally spaced. Look at a graph of the sine function.
 
  • #5
61
0
Then how should I go about solving this? The only idea I have is making this ito a triangle problem and finding out the angles and somehow relate it to distance and time
 
Last edited:
  • #6
OlderDan
Science Advisor
Homework Helper
3,021
2
Then how should I go about solving this? The only idea I have is making this ito a triangle problem and finding out the angles and somehow relate it to distance and time
At x = 0 your equation

y(x, t) = (0.35 m) sin[(1.25 rad/m)x + (108.2 rad/s)t]

reduces to

y(0, t) = (0.35 m) sin[(108.2 rad/s)t]

You need to find the first two positive values of t that make y(0, t) = 0.303 m. Substitute this into the above equation and solve for t, then take the difference between the first two values of t that satisfy the condition.
 
  • #7
61
0
Would just graphing the function and finding where .303 intersects with the function ok?
 
  • #8
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
9,621
6
Would just graphing the function and finding where .303 intersects with the function ok?
You could do that but it would be more accurate to solve the problem analytically, which in this case is not too difficult. So, we have;

[tex]0.35\sin(108.2(t)) = 0.303[/tex]

What do you think the next step would be?
 
  • #9
61
0
WEll I'm not particularly good with trig so i don't know what the second solution would be... you just divide by .35, take arcsin and then divide by 108.2 but I only get one of the answer, I don't know how to get the other one :P
 
  • #10
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
9,621
6
WEll I'm not particularly good with trig so i don't know what the second solution would be... you just divide by .35, take arcsin and then divide by 108.2 but I only get one of the answer, I don't know how to get the other one :P
HINT: Sketch the graph of y=sin(x), where is the next solution going to be relative to your particular solution?
 
  • #11
61
0
But sketching it won't give us an exact solution though. I mean I guess I see where it's suppose to be but I don't know how to find it exactly using equations and such.
 
  • #12
OlderDan
Science Advisor
Homework Helper
3,021
2
But sketching it won't give us an exact solution though. I mean I guess I see where it's suppose to be but I don't know how to find it exactly using equations and such.
There are exact relations between the trig functions that can be seen from the symmetry of the curves. For example, the cos(x) is 1 at x = 0, and by symmetry cos(-x) = cos(x). Another example is sin(x) = 0 at x = 0 and by symmetry sin(-x) = -sin(x). You can make a similar symmetry observation for the first two angles that satisfy your equation. One very precise value can be obtained with the calculator, and another can by found using this one value and the symmetry of the graph.
 

Related Threads on Need help with a few wave problems

Replies
2
Views
4K
Replies
2
Views
1K
  • Last Post
Replies
3
Views
7K
  • Last Post
Replies
9
Views
2K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
8
Views
1K
Replies
1
Views
1K
Replies
2
Views
7K
Replies
10
Views
1K
  • Last Post
Replies
6
Views
7K
Top