Waes: normal mode frequencies for 1 fixed extremity

[imphat]

Homework Statement

A String of length L has one of its extremities fixed and the other one loose.

A. What's the equation for the normal mode frequencies?
B. Draw a snapshot of the string for the 1st 3 normal modes

Homework Equations

wave equation

The Attempt at a Solution

My idea was to follow the same line of thinking for a string with both extremities fixed. Then we can assume that

y(x, t) = g(x).cos(wt + d) --(1)


and (1) must be solution of the wave equation, and after some math we get the general solution

A(x)'' + A.k^2 = 0, k = w/v --(2)


we know that A(0) = 0, since the x=0 extremity is fixed and the general solution for 2 is

A(x) = b.cos(kx) + c.sin(kx)

so, b = 0 and A(x) = c.sin(kx)

since A(x) != 0, we know that c != 0, but there's no other known condition in order to compute possible values for k


any ideas? and.. tbh i don't know if i can assume everything i did since 1 of the extremities is loose. help pls! :D

 

[anathema]

Your approach is on the right track, but there are a few things to consider when dealing with a string with one fixed and one loose end.

First, we can start by writing the wave equation for the string:

∂^2y/∂t^2 = v^2∂^2y/∂x^2

Where v is the velocity of the wave on the string. Since one end is fixed, we can impose the boundary condition y(0,t) = 0. This means that at x = 0, the displacement of the string is always 0, as you correctly stated.

For the other end, where the string is loose, we can impose a different boundary condition. Let's say that the string is free to move up and down at that end, meaning that the slope of the string at x = L is also 0. This can be written as ∂y/∂x = 0 at x = L.

Now, let's assume that the string is in its n-th normal mode, with a frequency of ωn. This means that the displacement of the string at any point x and time t can be written as:

y(x, t) = Ansin(knx)cos(ωnt)

Where An is the amplitude of the n-th normal mode and kn = nπ/L. This satisfies both boundary conditions we imposed earlier.

Substituting this into the wave equation, we get:

ωn^2Ansin(knx) = v^2k^2nAnsin(knx)

Dividing both sides by Ansin(knx), we get:

ωn^2 = v^2k^2n

Or, rearranging:

ωn = vkn

Which is the equation for the normal mode frequencies. So, the frequencies of the normal modes are multiples of the fundamental frequency, ω1 = vπ/L.

Now, for the snapshots of the first three normal modes, we can use the equation for y(x, t) given earlier. For the first normal mode (n = 1), we have:

y1(x, t) = A1sin(kx)cos(ω1t)

For the second normal mode (n = 2), we have:

y2(x, t) = A2sin(2kx)cos(ω2t)

And for the third normal mode (n = 3

 

[Moetasim]

Your approach is on the right track. However, since one extremity is fixed, the boundary conditions for the wave equation will be different. The boundary condition at x=0 will be A(0) = 0, as you have correctly stated. However, at the other extremity, x=L, the boundary condition will be A'(L) = 0, since the end is loose and there is no force acting on it.

Using these boundary conditions, you can solve for possible values of k and determine the normal mode frequencies. The first three normal modes will have k values of k1 = π/L, k2 = 2π/L, and k3 = 3π/L. These correspond to the first, second, and third normal modes, respectively.

To visualize these modes, you can plot the displacement of the string as a function of x for each mode. For the first mode, the displacement will be a simple sine curve with one node at x=L. For the second mode, there will be two nodes at x=L/2 and x=L. And for the third mode, there will be three nodes at x=L/3, x=2L/3, and x=L.

Overall, your approach is correct, but you just need to consider the different boundary conditions for a string with one fixed extremity.