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Energy of vibrating string
By Philip Ellison on Sunday, September 08, 2002 - 12:37 pm:

The midpoint of a violin string ocillates with amplitude 8mm and frequency 440Hz. It is suggested that the total energy of a string vibrating in this manner can be calculated using the equation:

E=1/2(m/2)v2, where m is the string's mass, and v is the maximum velocity of the midpoint of the string. Comment on this model.

The numbers are just for earlier parts of the question. However, could someone please help by explaining the derivation of the equation and describing its limitations. I have a feeling that this is very simple, I'm just missing the point entirely (I've never done any SHM before, so it doesn't really "make sense" in my head yet)! Thanks

By Andre Rzym on Tuesday, September 10, 2002 - 08:51 pm:

Assume, were the string at rest, that it would lie along the x-axis. Assume that any displacements are in the xy plane, and therefore the string is described at any time by y(x,t).

Assume the string has mass m per unit length and tension T (which remains constant).

Imagine the violin string as comprising a light (no mass, no resistance to bending) string with little 'beads' of mass at equal separations, dx. As dx->0 we get our violin string. Each bead has mass mdx.

Now imagine the bead at x. Resolve forces in the direction of the y-axis:

F=ma

where

a=acceleration=d2y/dt2 (derivatives here are partial as in all of what follows).
m=mass of bead=mdx
F=net force=sum of forces on either side of the bead
F=T.dy/dx|x+dx-T.dy/dx|x-dx
so
T.d2y/dx2=md2y/dt2

How do we sole this? Imagine a solution of the form
y(x,t)=A.f(x)g(t) (A a constant)
So
f''/f=g''/g.m/T
The LHS is a function of x only, the RHS is a function of y only, so they both must be constants.
Hence:
f(x)=sin(kx/Ö(T)+tk)
g(t)=cos(kt/Ö(m)+fk)

So
y(x,t)=Ak.sin(kx/Ö(T)+tk).cos(kt/Ö(m)+fk)

for any k,Ak,tk,fk

and the general solution is a linear combination of these over different k [i.e. an integral of the above function and an arbitrary function of k, integrated over k]

What of our violin string? Assume that one end is fixed at x=0, ie y(0,t)=0 "t.
=> tk=0 "k

Assume the other end is constrained at x=l, ie y(l,t)=0 "t.
=> k/l.Ö(T)=np for positive n

[we don't lose generality by ignoring -ve n since A can be -ve]

So

yn(x,t)=An.sin(npx/l).cos(npt.Ö(T/m)/l+fn)

Finally, assume the string is stationary at t=0 [i.e. we have plucked the string (in any shape) and release it at t=0] => fn=0 "n.

So our general solution (subject to the violin string assumptions) is

y=SnAn.sin(npx/l).cos(npt.Ö(T/m)/l)

It is not difficult to model this as a spreadsheet. Put t in a single cell (an input), put the An's (say 20 of them) as inputs. Assume some values for T etc (e.g. all 1) then compute the displacement of the string as a function of distance along the string. Divide the string into, say, 50 equal divisions and compute y for each. Then plot the result as a graph and watch how it changes for different A and as t advances. It is well worthwhile.

What values do we choose for the An's? The simple sinusoid I solve above is equivalent to setting Aa=a constant (to fit the peak velocity) and all the other A's are zero.

Nicholas's scenario of a triangular initial displacement is, I think (but it will be obvious if you do it on a spreadsheet, and you will see how the triangle loses its shape over time!!) given by
A1=1/12
A2=0
A3=1/32
A4=0
A5=1/52
etc

Unfortunately you need to be familiar with familiar Fourier analysis to see why.

Andre

By Philip Ellison on Wednesday, September 11, 2002 - 10:10 pm:

I originally tried a sinusiodal formula, but calculated the average velocity as 2v/pi. Could you please show me the correct calculations. Thanks very much.

By Andre Rzym on Thursday, September 12, 2002 - 08:24 am:

Philip,

If you are prepared to accept the assumptions:
a) The string is sinusoidal in shape and will continue to be so over time
b) The displacement of any point on the string is sinusoidal over time

then as I said above, the derivation is quite straightforward. To justify why these assumptions are valid requires the more complicated PDE approach that I went through above.

Assumption (a) implies that we can write:

y(x,t)=A.sin(ax+b)

since only the magnitude of the sine wave is assumed to vary, a & b must be constants.

At a given point of the string, sin(ax+b) is just a number, so to meet assumption (b) we need
A=B.sin(ct+d)
i.e.
y(x,t)=Bsin(ct+d).sin(ax+b)
where B,c,d,a,b are constants.
We don’t care about the position of the string at t=0 so we may as well choose d=0. Assume that the string is fixed at x=0 => b=0. Assume that the string is fixed at x=l => a=p/l.
Now
dy/dt=B.c.cos(ct).sin(px/l)
so the maximum speed at x=l / 2 (whence the sine term is 1) is
v=B.c
Now the total energy of the string over and above the energy when the string is stationary is switching between kinetic and potential energy. But when the centre is moving fastest, the string is straight (from the above equations). Therefore the total excess energy is just the kinetic energy at that time [when cos(ct)=1].

Between x,x+dx, the mass is just m/l.dx, the velocity is B.c.cost(ct).sin(px/l, which at the time of interest is just B.c.sin(px/l.

So total kinetic energy,

E=ò0l(1/2).( m/l.dx).(B2c2sin2(px/l)
E=1/2.m.v2ò0l(1/l).sin2( px/l).dx
E=1/2.(m/2).v2

Andre

By Andre Rzym on Thursday, September 12, 2002 - 08:28 am:

I think for other odd harmonics [a=3p/l etc] the formula will be the same but you would need to check

Andre

By Philip Ellison on Thursday, September 12, 2002 - 06:37 pm:

Could you please explain very carefully the reasons behind, and calculations necessary for the last bit (for "so total kinetic energy..."). I'm sorry to keep asking you for further help, but I'm being really dense and still don't get it! Thanks again for your patience!

By Andre Rzym on Friday, September 13, 2002 - 08:38 am:

I’ll try. I’ll also add numbers to the paragraphs so that you can identify places that still cause confusion

(i) Above the “so total kinetic energy …” we concluded that KE was greatest when the string was straight. We also know the maximum speed, v, in terms of two parameters that we do not know, B & c

(ii) So now (when KE is greatest and the string is straight) mentally chop the string into parts, each of length dx. Each piece is moving at different speeds. Each with its own kinetic energy. The total KE is the sum of the KE of each piece.

(iii) Focus on the piece between [x,x+dx] It’s KE is given by “1 / 2(mass)(speed)2”
What is “mass” here? The total mass of the string is m, the length is l so the mass per unit length is m/l so the mass of length dx is m/l.dx

(iv) What is “speed”? Well the string is straight when sin(ct)=0 => cos(ct)=+/-1 (ignore the minus sign since we are going to square it in a second).
We have
dy/dt=B.c.cos(ct)sin(px/l)
so a straight string implies
dy/dt=B.c.sin(px/l)
In other words, the maximum speed at any point varies with x (not too surprisingly).
(v) Now our “1 / 2(mass)(speed)2 looks like
(1 / 2)(m/l.dx)(B.c.sin(px/l))2=1 / 2.m/l.B2c2sin2(px/l).dx
We need to sum this over x=0 to l to get the total KE. But as dx->0 this sum tends to an integral, i.e.

E=ò0l1 / 2.m/l.B2c2sin2(px/l)dx=1 / 2.m/l.B2c2ò0lsin2( px/l)dx
(vi) We agreed that the maximum speed at the centre is given by
v=Bc
so substituting:
E=1 / 2.m/l.v2ò0lsin2( px/l)dx
(vii) It remains to evaluate this integral. First substitute y=px/l
E=1 / 2.m/l.v2ò0psin2y.dyl/p=1 / 2.mv2/pò0psin2ydy
(viii) How do we evaluate this integral? Well
cos(a+b)=cos(a)cos(b)-sin(a)sin(b)
cos(2y)=cos2(y) –sin2(y)
so
sin2y=1 / 2(1-cos(2y)
So E=1 / 2.m.v2.1/pò0 p1 / 2(1-cos(2y))dy
E=1 / 2.mv2.1/p.p / 2=1 / 2(m / 2)v2

Does this help? As an aside, the introduction is usually via an oscillating pendulum. Its easier!

Andre

By Philip Ellison on Friday, September 13, 2002 - 09:26 pm:

I finally get it... thanks so much! I didn't get the taking the Ek of each interval, and turning the equation into an integral idea from your first explanation, but your last post explained everything perfectly. Thanks again!