Musical Strings 2: UNIZOR.COM - Physics4Teens - Waves - Transversal Waves
Channel: Zor Shekhtman on UNIZOR Education
Date: 2021-10-17

UNIZOR.COM - Physics through a Looking Glass of Mathematician

Musical Strings 2

Let's examine the same model, as in the previous lecture, from the energy standpoint.
By lifting a point-mass initially by distance y(0)=a, 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 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
F↓↓(y) = −2·κ·y·[1 − (L/2) /√((L+l)/2)²+y²] = −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 y to y+dy of the infinitesimal length dy we can assume the constant force, as described above.
Therefore, the work we perform on that infinitesimal interval by going against the force F↓↓(y) is
dW = −F↓↓(y)·dy

We can now calculate the total work performed by displacing our point-mass by distance y(0)=a by integrating the above expression from y=0 to y=a.
W[0,a] = −∫[0,a]F↓↓(y)·dy = ∫[0,a]2·κ·y·[1−(L/2) /√((L+l)/2)²+y² ]·dy =
= ∫[0,a]2·κ·y·[1−L/√(L+l)²+4y² ]·dy

This integral can be calculated.
The answer is
W[0,a] = 2·κ·[y²/2 − (L/4)·√(L+l)²+4y² ] |[0,a] =
= 2·κ·[a²/2 − (L/4)·√(L+l)²+4a² + L·(L+l)/4] =
= κ·[a² − (L/2)·√(L+l)²+4a² + L·(L+l)/2]

Just to check, if a=0, the work performed is
W[0,0] = κ·[0² − (L/2)·√(L+l)²+0² + L·(L+l)/2] = 0
as it is supposed to be.

Analysis of the potential energy W(a) as a function of the initial displacement a shows that it resembles the parabola with a graph like this:

So, potential energy is always positive, as it should be, and grows with initial displacement a growing to positive or negative direction (up or down on a picture of a point-mass between two springs above).

At any displacement point y from the neutral position a point-mass m has certain potential and kinetic energy.
Its potential energy W(y), as a function of displacement y, looks exactly the same as the one we calculated at the initial position, when y=a, which yields
W(y) = κ·[y² − (L/2)·√(L+l)²+4y² + L·(L+l)/2]

Obviously, this formula is valid only for −a ≤ y ≤ a.
The difference between W[0,a] and W(y) is the kinetic energy our point-mass possesses because it moves towards the neutral position, that is
K(y)=m·(y')²/2.

Therefore, we have an equation, which is a differential equation for a displacement y(t) as a function of time t (in many cases we will drop (t) for brevity).
W(y) + K(y) = W[0,a]

The differential equation for displacement y(t) is
κ·[y² − (L/2)·√L+l)²+4y² + L·(L+l)/2] + m·(y')²/2 =
= κ·[a² − (L/2)·√(L+l)²+4a² + L·(L+l)/2]

Interestingly, though not surprisingly, this equation is not really different from the one derived in the previous lecture.
Take a derivative from both sides of an equation above, and you will transform it into the one derived in the previous lecture.