Determine max speed of point on string

  • Thread starter johnj7
  • Start date
  • #1
27
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Homework Statement


A 2.5 m long string that has a mass of 0.10 kg is fixed at both ends and is under a tension 30 N. (c) Determine the maximum speed of a point on the middle of the string.


Homework Equations


2 sinusoidal waves having same amplitude, frequency and wavelength / superposition

y(x, t) = 2Asin(kx)cos(wt)


The Attempt at a Solution


I have no idea as to what to do to approach this. The point on the string is moving in simple harmonic motion, which can be modeled by y(x) = Asin(wt) and v(x) = Awcos(wt)
so maximum velocity would be Aw. Is this right? how would you find A? I'm probably approaching this wrong. Could someone give me a hint?

For parts (a) we had to solve for the speed of the waves on string, which was
v = (T/u)^(1/2) = (30/0.04)^(1/2) = 27.39 m/s
for part (b) the question was "when the nth harmonic is excited, there is a node 0.50 m from one end. what is n?
wavelength = 2L/n
distance between nodes = wavelength / 2 = 0.5
wavelength = 1 m
1 = 2(2.5) / n
n = 5, 5th harmonic

I included parts (a) and (b) to show that I did think about this problem, I just got stuck with C. any help would be appreciated
 

Answers and Replies

  • #2
45
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The equation you gave,
y(x,t) = 2Asin(kx)cos(wt)
is an equation of two independent variables.

Every point on the string is given by only a single parameter, x.

To find the velocity of any given point at any given time, with respect to what variable do you have to differentiate y(x,t) by?
 
  • #3
27
0
right, so i need to differentiate y(x,t) with respect to t by keeping x constant
to get
v = -2Asin(kx)sin(wt)w
but then how do i get Amplitude?
 

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