*Notes to a video lecture on http://www.unizor.com*

__Musical Strings 2__

In the previous lecture we considered the model of a musical string as a point-mass

*between two identical springs of the length*

**m***each (where*

**(L+l)/2***is a neutral length of the string we are modeling and*

**L***is a stretch to create initial tension) with a coefficient of elasticity*

**l***.*

**k**Plucking a string is, therefore, modeled as an initial displacement of a point-mass vertically by

*.*

**y(0)=a**The result of the analysis of forces acting on a point-mass between these springs, when it performs transverse oscillations, was a differential equation

*[*

**y"(t) + 2·(κ/m)·y(t)·**

··

*]*

**1−L/√(L+l)²+4y²(t)**

**= 0**Let's examine this model from the energy standpoint.

By lifting a point-mass initially by distance

*, we perform some work against elastic forces of two springs. The forces are variable and depend on how much we stretch the springs. Let's take coordinate*

**y(0)=a***as our argument. Then, as explained in the previous lecture, the forces of two springs acting along the Y-axis against our motion total to*

**y***[*

**F**

·_{↓↓}(y) = −2·κ·y··

*]*

**1 − (L/2) /√((L+l)/2)²+y²***[*

**=**

= −2·κ·y·= −2·κ·y·

*]*

**1 − L /√(L+l)²+4y²**Incidentally, it's easy to see that an expression in [...] is always positive, therefore the sign of the force is always opposite to the sign of a displacement

*.*

**y**On an interval from

*to*

**y***of the infinitesimal length*

**y+**d**y***d*we can assume the constant force, as described above.

**y**Therefore, the work we perform on that infinitesimal interval by going

__against__the force

*is*

**F**_{↓↓}(y)*d*

**W = −F**d_{↓↓}(y)·**y**We can now calculate the total work performed by displacing our point-mass by distance

*by integrating the above expression from*

**y(0)=a***to*

**y=0***.*

**y=a**

**W**

_{[0,a]}*= −∫*_{[0,a]}*F*_{↓↓}(y)·*d*

**y =**

= ∫= ∫

**[**

_{[0,a]}*2·κ·y·*

··

*]*

**1−(L/2) /√((L+l)/2)²+y²**

**·**d**y =**

= ∫= ∫

**[**

_{[0,a]}*2·κ·y·*

··

*]*

**1−L/√(L+l)²+4y²**

**·**d**y**This integral can be calculated.

The answer is

**W****[**

_{[0,a]}*= 2·κ·**]*

**y²/2 −**

− (L/4)·√(L+l)²+4y²− (L/4)·√(L+l)²+4y²

*|*

**[**

_{[0,a]}*=*

= 2·κ·= 2·κ·

*]*

**a²/2 −**

− (L/4)·√(L+l)²+4a² +

+ L·(L+l)/4− (L/4)·√(L+l)²+4a² +

+ L·(L+l)/4

*[*

**=**

= κ·= κ·

*]*

**a² − (L/2)·√(L+l)²+4a² +**

+ L·(L+l)/2+ L·(L+l)/2

Just to check, if

*, the work performed is*

**a=0**

**W****[**

_{[0,0]}*=*

= κ·= κ·

*]*

**0² − (L/2)·√(L+l)²+0² +**

+ L·(L+l)/2+ L·(L+l)/2

**= 0**as it is supposed to be.

Analysis of the potential energy

*as a function of the initial displacement*

**W(a)***shows that it resembles the parabola with a graph like this:*

**a**So, potential energy is always positive, as it should be, and grows with initial displacement

*growing to positive or negative direction (up or down on a picture of a point-mass between two springs above).*

**a**At any displacement point

*from the neutral position a point-mass*

**y***has certain potential and kinetic energy.*

**m**Its potential energy

*, as a function of displacement*

**W(y)***, looks exactly the same as the one we calculated at the initial position, when*

**y***, which yields*

**y=a***[*

**W(y) =**

= κ·= κ·

*]*

**y² − (L/2)·√(L+l)²+4y² +**

+ L·(L+l)/2+ L·(L+l)/2

Obviously, this formula is valid only for

*.*

**−a ≤ y ≤ a**The difference between

**and**

*W*_{[0,a]}*is the kinetic energy our point-mass possesses because it moves towards the neutral position, that is*

**W(y)***.*

**K(y)=m·(y')²/2**Therefore, we have an equation, which is a differential equation for a displacement

*as a function of time*

**y(t)***(in many cases we will drop*

**t***for brevity).*

**(t)**

*W(y) + K(y) = W*_{[0,a]}The differential equation for displacement

*is*

**y(t)***[*

**κ·***]*

**y² − (L/2)·√L+l)²+4y² +**

+ L·(L+l)/2+ L·(L+l)/2

*[*

**+ m·(y')²/2 =**

= κ·= κ·

*]*

**a² − (L/2)·√(L+l)²+4a² +**

+ L·(L+l)/2+ L·(L+l)/2

Simplifying...

**y² − (L/2)·√(L+l)²+4y² +**

+ (m/κ)·(y')²/2 =

= a² − (L/2)·√(L+l)²+4a²+ (m/κ)·(y')²/2 =

= a² − (L/2)·√(L+l)²+4a²

or

**(m/2κ)·(y')² + y² −**

− (L/2)·√(L+l)² + 4y² =

= a² − (L/2)·√(L+l)²+4a²− (L/2)·√(L+l)² + 4y² =

= a² − (L/2)·√(L+l)²+4a²

This is a different differential equation that describes the transverse oscillations of a point-mass between two springs than that described in the previous lecture.

It contains only the first derivative of a displacement

*.*

**y(t)**Interestingly, though not surprisingly, this equation is not really different.

If two functions are equal, their derivatives are also equal. Take a derivative from both sides of an equation above:

**(m/2κ)·2y'·y" + 2y·y' −**

− (L/2)·8y·y' /2√(L+l)² + 4y² =

= 0− (L/2)·8y·y' /2√(L+l)² + 4y² =

= 0

or

**(m/2κ)·y'·y" + y·y' −**

− L·y·y' /√(L+l)² + 4y² =

= 0− L·y·y' /√(L+l)² + 4y² =

= 0

Now we can cancel

*getting*

**y'(t)**

**(m/2κ)·y" + y −**

− L·y /√(L+l)² + 4y² = 0− L·y /√(L+l)² + 4y² = 0

or

*[*

**(m/2κ)·y" + y·**

··

*]*

**1 − L /√(L+l)² + 4y²**

**= 0**or

*[*

**y" + (2κ/m)·y·**

··

*]*

**1 − L /√(L+l)² + 4y²**

**= 0**which is exactly the same as the equation derived in the previous lecture.

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