# How to find the amplitude of oscillations of a string with 5 beads?

• Redwaves
In summary: I am still puzzled about p. Looks to me like one of your equations in post 1 may have the p in the wrong place.I was thinking maybe I have to find all the eigenvector then get the transpose of the matrix to get the eigenvector for the modes.However, I think I have to find a way with the vector ##(3,0,\sqrt{3},2,2)##That link certainly clarifies matters.
Redwaves
Homework Statement
Find the amplitude of the modes 2 and 3 for string with 5 beads.
The string is fixed at each extremity.
The string is at rest and I have a displacement vector ( I'm not sure what that mean) $$(3,0,\sqrt{3},2,2)$$
Relevant Equations
$$A_n = sin(\kappa p)$$
$$A_n = cos(\kappa p)$$
Hi,
First of all, I'm wondering if a beaded string is the right term?

I have to find the amplitude of the modes 2 and 3 for a string with 5 beads.

In my book I have $$A_n = sin(\kappa p)$$ or $$A_n = cos(\kappa p)$$ it depends if the string is fixed or not I guess. where $$\kappa = \frac{n\pi p}{N+1}$$

Can I use this formula? It seems too easy.

robphy said:
I think you need to formulate the problem more clearly and completely for the reader and possibly for yourself.
Are you studying a set of coupled oscillators like
Or something like
http://farside.ph.utexas.edu/teaching/315/Waveshtml/node23.html ?
First oscillator, second link.
We can see there are 5 beads so 5 modes. Basically, I have to find the amplitudes for the second and third mode.
I'm so confused I have a hard time to formulate a question correctly.
I though I could simple use the formula above.

Redwaves said:
First oscillator, second link.
We can see there are 5 beads so 5 modes. Basically, I have to find the amplitudes for the second and third mode.
I'm so confused I have a hard time to formulate a question correctly.
I though I could simple use the formula above.
It is a bad idea to apply equations blindly without checking that they are appropriate. I cannot judge because you have not said what ##\kappa, p## represent.
The vector given appears to be the initial displacements, so you need to be using that somewhere.

haruspex said:
It is a bad idea to apply equations blindly without checking that they are appropriate. I cannot judge because you have not said what ##\kappa, p## represent.
The vector given appears to be the initial displacements, so you need to be using that somewhere.
Is ##\kappa## the different amplitudes for the modes?

I think ##
A_n = sin(\kappa p)
##
Means the amplitude of a beads in a specific mode,right?

##A'_{n,p} = sin(\frac{n \pi p}{N+1})##
should give the amplitude A' of the beads in the mode n. However, what's exactly the amplitude of the modes?

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Redwaves said:
I think ##A_n = sin(\kappa p)##
Means the amplitude of a beads in a specific mode,
Yes, that was my guess too.
Redwaves said:
Is ##\kappa## the different amplitudes for the modes?
How can it be if the ##A_n ## are the amplitudes?
Since you have an equation expressing ##\kappa## as a function of p, the more fundamental question is the meaning of p. But seeing ##\kappa## gets recombined with p in the amplitude equation (resulting in a ##p^2## contribution), it would seem ##\kappa## has some natural meaning of its own.

Surely the place from which you are reading these equations defines its variables?

##
\kappa = \frac{n\pi p}{N+1}
##

But I have no idea what p means.

Even here, http://farside.ph.utexas.edu/teaching/315/Waveshtml/node23.html
I don't see how to find the amplitudes of each mode.

I was thinking maybe I have to find all the eigenvector then get the transpose of the matrix to get the eigenvector for the modes.
However, I think I have to find a way with the vector ##(3,0,\sqrt{3},2,2)##

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Redwaves said:
##
\kappa = \frac{n\pi p}{N+1}
##

But I have no idea what p means.

Even here, http://farside.ph.utexas.edu/teaching/315/Waveshtml/node23.html
I don't see how to find the amplitudes of each mode.

I was thinking maybe I have to find all the eigenvector then get the transpose of the matrix to get the eigenvector for the modes.
However, I think I have to find a way with the vector ##(3,0,\sqrt{3},2,2)##
That link certainly clarifies matters.
We are dealing with transverse oscillations; I had guessed longitudinal.
In the zero displacement position, there is already substantial tension in the string. So much so that we can ignore the extra tension resulting from the displacements. (The text at the link fails to make that point correctly.)

The text shows that the general pattern of displacements at time t is given by ##y_i=A\sin(kx_i)\cos(\omega t-\phi)##, where ##k=\frac{n\pi}{(N+1)a}##.
a is the horizontal separation, so (N+1)a is the length of the string.
n is the mode number; it is the number of half wavelengths along the string. See the diagrams at the link for examples of different mode numbers.

I am still puzzled about p. Looks to me like one of your equations in post 1 may have the p in the wrong place.

(Edit: para deleted here, replaced by post #10.)

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I can use the starting vector to find the constants in all modes? I'm not sure to understand why?

i = p in the link (243)
I think this is the position of the beads that we choose.

"The first constraint is automatically satisfied, because
"
Since the string if fixed I have ##x_0 = 0 ## in all modes?

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Redwaves said:
i = p in the link (243)
That makes sense for $$\kappa = \frac{n\pi p}{N+1}$$, though I dislike it because it hides the dependency of ##\kappa## on i (=p).
But these two I cannot fathom:
$$A_n = sin(\kappa p)$$
$$A_n = cos(\kappa p)$$
As stated in the text following (244), ##A_n, \phi_n## are arbitrary constants determined by the initial conditions, i.e., by the given position vector.
Using kappa, equation (243) becomes
##y_{n,i}(t)=A_n\sin(\kappa)\cos(...)##.

##x_0## is the position of an imaginary bead at one fixed end of the string, so ##x_0=y_0=0##. Likewise, ##x_{N+1}## is the other end of the string, so ##y_{N+1}=0##.
Redwaves said:
I can use the starting vector to find the constants in all modes?
My last paragraph in post #8 was wrong.
In general, the motion is the sum of any number of normal modes, each with its own amplitude. Just like Fourier analysis. The given initial position (and using the fact that the velocities are all zero at that instant) is enough to determine all amplitudes, but you are only asked for those corresponding to n=2 and n=3.

##
y_{n,i}(t)=A_n\sin(\kappa)\cos(\omega_n t - \phi_n)
##

then
##V_{n,i}(t) = -\omega_nA_n sin(\omega_n t - \phi_n)##

##v=0## then ##y_{n,i}## can be 2
t = 0 for both v and y ?
is it correct?

When you say I should use the vector, it is the modulus, the sum of
##(3,0,\sqrt{3},2,2)## or just any of the position?

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Redwaves said:
##
y_{n,i}(t)=A_n\sin(\kappa)\cos(\omega_n t - \phi_n)
##

then
##V_{n,i}(t) = -\omega_nA_n sin(\omega_n t - \phi_n)##

##v=0## then ##y_{n,i}## can be 2
t = 0 for both v and y ?
is it correct?

When you say I should use the vector, it is the modulus, the sum of
##(3,0,\sqrt{3},2,2)## or just any of the position?
As I wrote, the actual movement is a sum of modes:
##y_{i}(t)=\Sigma_nA_n\sin(\kappa)\cos(\omega_n t - \phi_n)##.
And I don't like using this kappa. Better as
##y_{i}(t)=\Sigma_nA_n\sin(\frac{n\pi i}{N+1})\cos(\omega_n t - \phi_n)##.
Then differentiating:
##v_{i}(t)=\Sigma_n-\omega_nA_n\sin(\frac{n\pi i}{N+1})\sin(\omega_n t - \phi_n)##.
(You left out the ##\sin(\kappa)## term.)
At t=0:
##0=\Sigma_n-\omega_nA_n\sin(\frac{n\pi i}{N+1})\sin(- \phi_n)## for all i.
Plus, the five equations for the initial positions.

haruspex said:
As I wrote, the actual movement is a sum of modes:
##y_{i}(t)=\Sigma_nA_n\sin(\kappa)\cos(\omega_n t - \phi_n)##.
I see, it is always like that?
The movement of the string is always the sum of modes?

So I have 5 equations for the position and 5 for the movement?
Basically,5 equetions (5 * i) for each mode. So a total of 25 equations for the velocity. I must made an error.
is ##\phi_n## the angle of the fixed position at rest?

Edit:
I have 5 equations for v = 0 and 5 equations for y(t) = 0 (the position for the second bead)
Thus, I have 10 equations for 1 bead.
For n = 2 I still have 3 unknown constants (##\omega_2 , A_2, \phi_2##)with only 2 equations.
However, As you said ##\phi## should be determined by the initial condition, but isn't ##\phi## the angle with the horizontal axis between 2 beads?

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Redwaves said:
The movement of the string is always the sum of modes?
We can think of a solution as a vector ##(y_i(t))=\vec y(t)##. Given any two solutions, ##\vec y_1(t), \vec y_2(t)##, any linear sum ##a_1\vec y_1+a_2\vec y_2## is also a solution.
The 'normal mode' solutions are those of the form in (236), but the general solution is any linear combination of these.
Redwaves said:
I have 5 equations for v = 0 and 5 equations for y(t) = 0 (the position for the second bead)
You have 5 equations of the form ##v_i(0)=0## and 5 of the form ##y_i(0)=..##.

haruspex said:
You have 5 equations of the form ##v_i(0)=0## and 5 of the form ##y_i(0)=..##.
Here's what I got.
For v(t)
##0 = -\omega_1 A_1 sin(\frac{\pi}{6}) sin(-\phi_1)##
##0 = -\omega_2 A_2 sin(\frac{2 \pi}{6}) sin(-\phi_2)##
##0 = -\omega_3 A_3 sin(\frac{3 \pi}{6}) sin(-\phi_3)##
##0 = -\omega_4 A_4 sin(\frac{4 \pi}{6}) sin(-\phi_4)##
##0 = -\omega_5 A_5 sin(\frac{5 \pi}{6}) sin(-\phi_5)##

For ##y_2(0)## Where the position for this bead is 0
##0 = A_1 sin(\frac{\pi}{6}) cos(-\phi_1)##
##0 = A_2 sin(\frac{2 \pi}{6}) cos(-\phi_2)##
##0 = A_3 sin(\frac{3 \pi}{6}) cos(-\phi_3)##
##0 = A_4 sin(\frac{4 \pi}{6}) cos(-\phi_4)##
##0 = A_5 sin(\frac{5 \pi}{6}) cos(-\phi_5)##

Redwaves said:
Here's what I got.
For v(t)
##0 = -\omega_1 A_1 sin(\frac{\pi}{6}) sin(-\phi_1)##
##0 = -\omega_2 A_2 sin(\frac{2 \pi}{6}) sin(-\phi_2)##
##0 = -\omega_3 A_3 sin(\frac{3 \pi}{6}) sin(-\phi_3)##
##0 = -\omega_4 A_4 sin(\frac{4 \pi}{6}) sin(-\phi_4)##
##0 = -\omega_5 A_5 sin(\frac{5 \pi}{6}) sin(-\phi_5)##

For ##y_2(0)## Where the position for this bead is 0
##0 = A_1 sin(\frac{\pi}{6}) cos(-\phi_1)##
##0 = A_2 sin(\frac{2 \pi}{6}) cos(-\phi_2)##
##0 = A_3 sin(\frac{3 \pi}{6}) cos(-\phi_3)##
##0 = A_4 sin(\frac{4 \pi}{6}) cos(-\phi_4)##
##0 = A_5 sin(\frac{5 \pi}{6}) cos(-\phi_5)##
No, you are still confused.
Each equation should have a sum over an unlimited number of modes (n) for a single value of i.
Use my second and third equations in post #12, substituting the five different values for i.

Yeah, I'm really confused with that part. My professor has skipped the entire chapter and without example in class it is difficult for me to understand.
at t = 0
##3=\Sigma_nA_n\sin(\frac{n\pi}{N+1})\cos( - \phi_n)##
##0=\Sigma_nA_n\sin(\frac{2 n\pi}{N+1})\cos( - \phi_n)##
##\sqrt{3}=\Sigma_nA_n\sin(\frac{3 n\pi}{N+1})\cos(- \phi_n)##
##2=\Sigma_nA_n\sin(\frac{4 n\pi}{N+1})\cos( - \phi_n)##
##2=\Sigma_nA_n\sin(\frac{5 n\pi}{N+1})\cos( - \phi_n)##

Each position is the sum of all modes.

##0=\Sigma_n-\omega_nA_n\sin(\frac{n\pi }{N+1})\sin( - \phi_n)##
##0=\Sigma_n-\omega_nA_n\sin(\frac{2 n\pi }{N+1})\sin( - \phi_n)##
##0=\Sigma_n-\omega_nA_n\sin(\frac{3 n\pi }{N+1})\sin( - \phi_n)##
##0=\Sigma_n-\omega_nA_n\sin(\frac{4 n\pi }{N+1})\sin( - \phi_n)##
##0=\Sigma_n-\omega_nA_n\sin(\frac{5 n\pi }{N+1})\sin(- \phi_n)##

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Redwaves said:
Yeah, I'm really confused with that part. My professor has skipped the entire chapter and without example in class it is difficult for me to understand.
at t = 0
##3=\Sigma_nA_n\sin(\frac{n\pi}{N+1})\cos( - \phi_n)##
##0=\Sigma_nA_n\sin(\frac{2 n\pi}{N+1})\cos( - \phi_n)##
##\sqrt{3}=\Sigma_nA_n\sin(\frac{3 n\pi}{N+1})\cos(- \phi_n)##
##2=\Sigma_nA_n\sin(\frac{4 n\pi}{N+1})\cos( - \phi_n)##
##2=\Sigma_nA_n\sin(\frac{5 n\pi}{N+1})\cos( - \phi_n)##

Each position is the sum of all modes.

##0=\Sigma_n-\omega_nA_n\sin(\frac{n\pi }{N+1})\sin( - \phi_n)##
##0=\Sigma_n-\omega_nA_n\sin(\frac{2 n\pi }{N+1})\sin( - \phi_n)##
##0=\Sigma_n-\omega_nA_n\sin(\frac{3 n\pi }{N+1})\sin( - \phi_n)##
##0=\Sigma_n-\omega_nA_n\sin(\frac{4 n\pi }{N+1})\sin( - \phi_n)##
##0=\Sigma_n-\omega_nA_n\sin(\frac{5 n\pi }{N+1})\sin(- \phi_n)##
Yes, that looks right.
Now, you should find that there is only a finite set of values of n that are worth considering. After that, the sin functions start repeating.
Also, there is an obvious set of values for the phases that makes all the velocities zero.
That should leave you with five equations and five unknowns.

##\phi = n\pi## for n in R
The five unknowns are ##A_n## which is the amplitude of all modes together.
I have 5 equations like this
##3 = A_1 sin(\frac{\pi}{6}) + A_2 sin(\frac{2\pi}{6}) + A_3 sin(\frac{3\pi}{6}) + A_4 sin(\frac{4\pi}{6}) + A_5 sin(\frac{5\pi}{6})##

Redwaves said:
##\phi = n\pi## for n in R
The five unknowns are ##A_n## which is the amplitude of all modes together.
I have 5 equations like this
##3 = A_1 sin(\frac{\pi}{6}) + A_2 sin(\frac{2\pi}{6}) + A_3 sin(\frac{3\pi}{6}) + A_4 sin(\frac{4\pi}{6}) + A_5 sin(\frac{5\pi}{6})##
Each An is the amplitude of one mode.
Not sure if that's shat you meant.
The sample equation looks right.

haruspex said:
Each An is the amplitude of one mode.
Not sure if that's shat you meant.
The sample equation looks right.
Yes that is what I meant.
However, it's a mess to find the amplitudes.

Redwaves said:
Yes that is what I meant.
However, it's a mess to find the amplitudes.
It's not that bad.
Write it out in matrix form and do row additions to eliminate terms. I got A3 in a few steps.

I get this matrix, but I should get ##-\frac{1}{2}## for the amplitude in the second mode, order by frequency. but none of my amplitudes has this value.
##
\begin{pmatrix}
-1 & \sqrt{3} & -2 & \sqrt{3} & -1 & 6\\
1 & 1 & 0 & -1 & -1 & 0\\
1 & 0 & -1 & 0 & 1 & \sqrt{3}\\
1 & -1 & 0 & 1 & -1 & \frac{4}{\sqrt{3}}\\
\frac{1}{\sqrt{3}} &-1 & \frac{2}{\sqrt{3}} & -1 & \frac{1}{\sqrt{3}} & \frac{4}{\sqrt{3}}\\
\end{pmatrix}
##

Redwaves said:
I get this matrix, but I should get ##-\frac{1}{2}## for the amplitude in the second mode, order by frequency. but none of my amplitudes has this value.
##
\begin{pmatrix}
-1 & \sqrt{3} & -2 & \sqrt{3} & -1 & 6\\
1 & 1 & 0 & -1 & -1 & 0\\
1 & 0 & -1 & 0 & 1 & \sqrt{3}\\
1 & -1 & 0 & 1 & -1 & \frac{4}{\sqrt{3}}\\
\frac{1}{\sqrt{3}} &-1 & \frac{2}{\sqrt{3}} & -1 & \frac{1}{\sqrt{3}} & \frac{4}{\sqrt{3}}\\
\end{pmatrix}
##
I get the same matrix except for the top line, where I have all positive signs.

haruspex said:
I get the same matrix except for the top line, where I have all positive signs.
I get the same, I made a mistake when I type it here, but, I don't get the right answer.

Redwaves said:
I get the same, I made a mistake when I type it here, but, I don't get the right answer.
I get A1 etc as
##\frac{7+8\sqrt 3}{12}, -\frac {5\sqrt 3}{12}, \frac{14-4\sqrt 3}{12}, \frac {3\sqrt 3}{12}, \frac 7{12}##.

Redwaves said:
I have 5 equations like this
##3 = A_1 sin(\frac{\pi}{6}) + A_2 sin(\frac{2\pi}{6}) + A_3 sin(\frac{3\pi}{6}) + A_4 sin(\frac{4\pi}{6}) + A_5 sin(\frac{5\pi}{6})##
Yes. Or in compact notation, $$y_p = \sum_{n = 1}^{N} A_n \sin\left(\frac{n\pi p}{N+1}\right)\,\,\,\,\,\,\,\, (\rm eq. 1)$$ where ##N## is the number of beads and ##y_p## is the initial displacement of the ##p^{\rm th}## bead.

Here's an alternate way to find the ##A_n##. The normal modes satisfy an orthogonality condition:
If ##N##, ##m##, and ##n## are any integers, then $$\sum_{p = 1}^{N} \sin\left(\frac{m\pi p}{N+1}\right) \sin\left(\frac{n\pi p}{N+1}\right) = \frac{N+1}{2} \delta_{n,m}\,\,\,\,\,\,\,\, (\rm eq. 2)$$
Multiply both sides of (eq. 1) by ##\sin\left(\frac{m\pi p}{N+1}\right) ## and sum over ##p##. Then use (eq. 2).

Delta2 and haruspex
haruspex said:
I get A1 etc as
##\frac{7+8\sqrt 3}{12}, -\frac {5\sqrt 3}{12}, \frac{14-4\sqrt 3}{12}, \frac {3\sqrt 3}{12}, \frac 7{12}##.
So I should have made an error somewhere.
Since, the right answer are ##A_2 = -\frac{1}{2}, A_3 = \frac{5-\sqrt{3}}{\sqrt{3}}## for the amplitudes order by frequency.

TSny said:
Yes. Or in compact notation, $$y_p = \sum_{n = 1}^{N} A_n \sin\left(\frac{n\pi p}{N+1}\right)\,\,\,\,\,\,\,\, (\rm eq. 1)$$ where ##N## is the number of beads and ##y_p## is the initial displacement of the ##p^{\rm th}## bead.

Here's an alternate way to find the ##A_n##. The normal modes satisfy an orthogonality condition:
If ##N##, ##m##, and ##n## are any integers, then $$\sum_{p = 1}^{N} \sin\left(\frac{m\pi p}{N+1}\right) \sin\left(\frac{n\pi p}{N+1}\right) = \frac{N+1}{2} \delta_{n,m}\,\,\,\,\,\,\,\, (\rm eq. 2)$$
Multiply both sides of (eq. 1) by ##\sin\left(\frac{m\pi p}{N+1}\right) ## and sum over ##p##. Then use (eq. 2).
I'm not sure to understand where the eq 2 came from.

I thought I should be able to explain that equation but I got stuck.
For convenience, I've used N instead of N+1.

##4\sin(\pi p\frac mN) \sin(\pi p\frac nN) =(e^{i\pi p\frac mN}-e^{-i\pi p\frac mN})(e^{i\pi p\frac nN}-e^{-i\pi p\frac nN})##
##=e^{i\pi (p\frac mN+p\frac nN)}+e^{-i\pi (p\frac mN+p\frac nN)}-e^{i\pi (p\frac mN-p\frac nN)}-e^{i\pi (p\frac nN-p\frac mN)}##
Summing the geometric series for the first of those four terms:
##\Sigma_{p=0}^{N-1}e^{i\pi (p\frac {m+n}N)}=\frac{1-e^{i\pi (N\frac {m+n}N)}}{1-e^{i\pi (\frac {m+n}N)}}=\frac{1-e^{i\pi ( m+n)}}{1-e^{i\pi (\frac {m+n}N)}}##
If m+n is even then the numerator vanishes, but it's not going to if m+n is odd.

The same applies to the sum of the m-n terms except for the special case of m=n:
##\Sigma_{p=0}^{N-1}-e^{i\pi (p\frac nN-p\frac nN)}=-N##

haruspex said:
Summing the geometric series for the first of those four terms:
##\Sigma_{p=0}^{N-1}e^{i\pi (p\frac {m+n}N)}=\frac{1-e^{i\pi (N\frac {m+n}N)}}{1-e^{i\pi (\frac {m+n}N)}}=\frac{1-e^{i\pi ( m+n)}}{1-e^{i\pi (\frac {m+n}N)}}##
If m+n is even then the numerator vanishes, but it's not going to if m+n is odd.
For m+n odd, the numerator in your final expression is 2.

So, ##S1 \equiv \Sigma_{p=0}^{N-1}e^{i\pi (p\frac {m+n}N)} = \frac{2}{1-e^{i\pi (\frac {m+n}N)}} ##

Likewise, ##S2 \equiv \Sigma_{p=0}^{N-1}e^{-i\pi (p\frac {m+n}N)} = \frac{2}{1-e^{-i\pi (\frac {m+n}N)}}##.

Then, ##S1+S2 = 2##.

If m+n is odd, then m-n is also odd.

So, ##S3 + S4 = 2##, where ##S3 \equiv \Sigma_{p=0}^{N-1}e^{i\pi (p\frac {m-n}N)} ## and ##S4 \equiv \Sigma_{p=0}^{N-1}e^{-i\pi (p\frac {m-n}N)} ##

Thus, overall, we get ##S1 + S2 - S3 - S4 = 0##.

Delta2 and haruspex
Redwaves said:
So I should have made an error somewhere.
Since, the right answer are ##A_2 = -\frac{1}{2}, A_3 = \frac{5-\sqrt{3}}{\sqrt{3}}## for the amplitudes order by frequency.
What did you get for ##A_2##?

TSny said:
What did you get for ##A_2##?
I tried your method, but I don't understand. we didn't use complex numbers.
I finally got an answer from my professor by email. However, I think I'm more confuse now.
He told me I should use the formula to find the eigenvectors of normal modes which is ##<n| = \frac{1}{\sqrt{\sum_N}}(sin(\frac{n\pi p}{N+1}, sin(\frac{n\pi p}{N+1},...)## Where N is the number of beads, p the index of the bead and n the mode.

However, I don't see how to know which mode is the second and third mode order by frequency.

I'm not sure, but I think the eigenvectors is the ratio of the different amplitude. thus, maybe the lowest amplitude is the higher frequency, but I'm just guessing.
As I type that I think what I said makes not sense since I only get the eigenvector for a single mode with are the ratio of different amplitude only in a specific mode. I'm rock solid stuck.

Edit:
Maybe it will sound stupid, but is the modulus of the eigenvector the amplitude?

Edit #2:
I think I have to find a relation between the eigenvector and the amplitude.

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The nth normal mode is described by the expression $$\sin\left(\frac{n\pi p}{N+1}\right)$$ This expression gives the relative amplitude of the pth bead in the nth normal mode. The initial displacements of the beads in your problem are (y1, y2, y3, y4, y5) = (3, 0, √3, 2, 2). These can be expressed as a superposition of the normal mode expressions $$y_p = \sum_{n = 1}^{N} A_n \sin\left(\frac{n\pi p}{N+1}\right) = \sum_{n = 1}^{5} A_n \sin\left(\frac{n\pi p}{6}\right)$$ The ##A_n## are the amplitudes of the normal modes. The normal modes obey an orthogonality relation $$\sum_{p = 1}^{N} \sin\left(\frac{m\pi p}{N+1}\right) \sin\left(\frac{n\pi p}{N+1}\right) = \frac{N+1}{2} \delta_{n,m}$$ or$$\sum_{p = 1}^{5} \sin\left(\frac{m\pi p}{6}\right) \sin\left(\frac{n\pi p}{6}\right) = 3 \delta_{n,m}$$ The occurrence of the ##3## on the right hand side of the last equation means that the normal mode expressions are not "normalized". You can see that if we choose the normal modes as $$\frac{1}{\sqrt{3}}\sin\left(\frac{n\pi p}{6}\right)$$ where we have included a "normalization factor" of ##\frac{1}{\sqrt{3}}##, then we have $$\sum_{p = 1}^{5} \left[ \frac{1}{\sqrt 3} \sin\left(\frac{m\pi p}{6}\right) \right] \left[ \frac{1}{\sqrt 3} \sin\left(\frac{n\pi p}{6}\right) \right] = \delta_{n,m}$$ With these normalized normal modes, the right hand side is just ##\delta_{n,m}## with a coefficient of 1. For the case where ##m = n##, $$\sum_{p = 1}^{5} \left[ \frac{1}{\sqrt 3} \sin\left(\frac{n \pi p}{6}\right) \right]^2 = 1 \,\,\,\,\,\,\,$$ for each normalized normal mode (##n = 1, 2, ..., 5)##.

Using these normalized modes, the initial displacements ##y_p## can be expressed as $$y_p = \sum_{n = 1}^{5} B_n \left[ \frac 1 {\sqrt 3} \sin\left(\frac{n\pi p}{6}\right) \right]$$ where I'm using ##B_n## to denote the amplitude of the nth normalized normal mode.

You are asked to find the values of ##B_n## for ##n = 2## and ##n=3##. To do this, multiply both sides of this equation by the mth normal mode ##\frac 1 {\sqrt 3} \sin\left(\frac{m\pi p}{6}\right)## and sum over the index ##p##. Using the orthogonality condition, you should end up with an expression for calculating ##B_m## in terms of the given values of ##y_p##.

Last edited:
Redwaves
That's what I tried, but I didn't get the right answer. For example, I have ##B_n = 0## for ##n = 2## and ##B_n = 3## for ##n = 3##

for ##n=3##
##\sqrt{3}=B_n[\frac{1}{\sqrt{3}}(1+0-1+0+1)] = B_n \frac{1}{\sqrt{3}}##
thus, if I multiply both side by ##\frac{1}{\sqrt{3}}##
I have ##1 = B_n(\frac{1}{3})##

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