Initial velocity of bird landing on horizontal wire modeled as spring

[zenterix]

Homework Statement

A massive bird lands on a taut light horizontal wire. You observe that after landing the bird oscillates up and down with approximately simple harmonic motion with period $T$.

You notice that during the initial oscillations the maximum height the bird reaches above its initial position at landing is equal to the distance below its landing position at which the bird finally comes to rest, when all oscillations have stopped.

Relevant Equations

In terms of only $T$ and $g$, the acceleration due to gravity, estimate the magnitude of the vertical velocity that the bird had at the instant of landing.

Here is a picture of the situation

Let's model the wire as acting like a massless spring with spring constant $k$.

Let's define our coordinate system to be a $y$-axis that has zero at the initial position of the wire (the landing position, and the equilibrium position of the spring).

At the bottom of the harmonic motion, velocity is zero and the gravitational force has same magnitude and opposite direction to the spring force.

$$mg\hat{j}-kh\hat{j}=0\implies h=\frac{mg}{k}$$

Let's define the gravitational potential energy to be zero at $h$. Then, the gravitational potential energy is

$$U_g(y)=mg(h-y)$$

and the spring potential energy is

$$U_s(y)=\frac{ky^2}{2}$$

Let's consider our system to be just the bird. There are two conservative external forces acting on the bird: the force of gravity and the spring force. Thus, mechanical energy is conserved and we can equate initial to final mechanical energies.

$$U_{g,i}+U_{s,i}+K_i=U_{g,f}+U_{s,f}+K_f$$

$$mgh+\frac{mv_i^2}{2}=\frac{kh^2}{2}\tag{1}$$

If we sub in $h=\frac{mg}{k}$ and solve for $v_i$ we obtain

$$v_i=\pm g\sqrt{-\frac{m}{k}}$$

and if we use $T=\frac{2\pi}{\omega}$ and $\omega=\sqrt{\frac{k}{m}}$ then we find that

$$v_i=\pm \frac{gT}{2\pi}i$$

This is a complex number. Now, I do not think this is correct. In fact, from the resources I am using it seems the initial velocity is

$$v_i=\frac{\sqrt{3}gT}{2\pi}$$

which is what would be obtained if in (1) we had $-mgh$ instead of $mgh$.

What (probably silly) mistake am I making here?

 

[kuruman]

Here is your silly mistake

At the bottom of the harmonic motion, velocity is zero and the gravitational force has same magnitude and opposite direction to the spring force.

The velocity is indeed zero but the acceleration is not. Zero velocity does not necessarily mean zero net force. The forces are balanced at the equilibrium point where the bird finally comes to rest.

That said, you can make your analysis simpler if you ignore gravity and treat the problem as a horizontal spring-mass system with the amplitude of oscillations measured from the equilibrium position of the bird.

 

[zenterix]

Okay, what if I disregard the assumption of simple harmonic motion and just consider two states: right before landing on the wire, and after the motion has been damped so that the bird is at rest on the wire?

Is the final state not one in which $mg\hat{j}-kh\hat{j}=0$?

 

[zenterix]

That said, you can make your analysis simpler if you ignore gravity and treat the problem as a horizontal spring-mass system with the amplitude of oscillations measured from the equilibrium position of the bird.

I think I do get this. The gravitational force on the bird is constant and simply generates an offset in the spring's equilibrium position. At position $$\frac{mg}{k}$$ the gravitational force and the spring force offset one another. Simple harmonic motion occurs about this position.

Changes in $g$ don't affect the frequency or period, but changes in $k$ do because it is the spring force that is generating the SHM (which originates from the homogeneous solution to the differential equation of motion). Gravity is present in the inhomogeneous term, which is just a constant external input that generates a constant term in the solution but not oscillations.

Ignoring gravity, as you suggest, means considering just the homogeneous equation of motion, which is what generates the oscillations.

 

[kuruman]

Is the final state not one in which $mg\hat{j}-kh\hat{j}=0$?

Yes that is the final state.

Simple harmonic motion occurs about this position.

Right. What is the amplitude of the motion? How is the amplitude related to the total energy of the oscillator before any significant amount is damped?

 

[zenterix]

I will do these calculations right now. I do have a question about the math involved, if I may.

Consider the following calculation.

This way of doing these calculations using complex exponentials is interesting and new to me, but I don't quite understand it yet. In particular, in the second line, why do we take the real part of that specific expression?

Up to this point, I would do these calculations in the following way

$$c\cos{\omega t}+d\sin{\omega t}$$

We search for an angle $\phi$ such that $c=A\cos{\phi}$ and $d=A\sin{\phi}$. This can be visualized as follows

Then

$$A=\sqrt{c^2+d^2}$$

$$\phi=\tan^{-1}{\frac{c}{d}}$$

and

$$c\cos{(\omega t)}+d\sin{(\omega t)}$$

$$=(A\cos{(\phi)})\cos{(\omega t)}+(A\sin{(\phi))\sin{(\omega t)}}$$

$$=A(\cos{(\omega t)}\cos{(\phi)}+\sin{(\omega t)}\sin{(\phi)})$$

$$=A\cos{(\omega t-\phi)}$$

I am asking about this because this calculation is needed to calculate the amplitude of the oscillation of the problem in the OP. I'd like to do it using complex exponentials.

Here is what I think

We have

$$c(e^{i\omega t}+e^{-i\omega t})/2-id(e^{i\omega t}-e^{-i\omega t})/2$$

and if we group these terms in a specific way

$$=\left (\frac{ce^{i\omega t}}{2}-id\frac{e^{i\omega t}}{2}\right )+\left (\frac{ce^{-i\omega t}}{2}+id\frac{e^{-i\omega t}}{2}\right )$$

Then it turns out that if we check the real part of either of the expressions in parentheses we get the original expression.

So, as far as I can tell, the answer to my question is simply that this is an algebraic truth. I guess my question originated from the fact that this step was merely presented without explanation in the book. Now that I checked I can see it is true.

 

[zenterix]

The equation of motion is

$$mg\hat{j}-ky\hat{j}=m\ddot{y}\hat{j}$$

$$\ddot{y}+\omega^2y=g$$

where $\omega^2=\sqrt{\frac{k}{m}}$.

The general solution is

$$y(t)=\frac{g}{\omega^2}+c_1\cos{\omega t}+c_2\sin{\omega t}$$

and so

$$\dot{y}(t)=-c_1\omega\sin{\omega t}+c_2\omega\cos{\omega t}$$

Our initial conditions are

$$y(0)=0$$

$$\dot{y}(0)=v_0$$

from which we obtain

$$c_1=-\frac{g}{\omega^2}$$

$$c_2=\frac{v_0}{\omega}$$

Thus

$$y(t)=\frac{g}{\omega^2}-\frac{g}{\omega^2}\cos{\omega t}+\frac{v_0}{\omega}\sin{\omega t}$$

$$=\frac{g}{\omega^2}+A\cos{(\omega t-\phi)}$$

where

$$A=\sqrt{\frac{g^2}{\omega^4}+\frac{v_0^2}{\omega^2}}$$

$$\phi=\tan^{-1}{\left (-\frac{g}{v_0\omega}\right )}$$

 

[zenterix]

In post #6 above I asked a question about manipulations of complex exponentials. I want to clear up my doubts in this post.

Consider the function

$$f(t)=a\cos{\omega t}+b\sin{\omega t}$$

and its derivative

$$f'(t)=-a\omega\sin{\omega t}+b\omega\cos{\omega t}$$

Right off the bat we can already determine that

$$a=f(0)$$

$$b=\frac{1}{\omega}\dot{f}(0)$$

and so

$$f(t)=f(0)\cos{\omega t}+\frac{1}{\omega}f'(0)\sin{\omega t}$$

What is special about this function is that it is the unique solution to the IVP

$$\ddot{f}(t)+\omega^2 f(t)=0$$

$$f(0)=a$$

$$f'(0)=b\omega$$

The question at this point is: why can we write $f$ in amplitude-phase form, why is it useful, and what does it mean?

The first way to think about this is quite mechanical in nature.

If we imagine that $a$ and $b$ are the coordinates in the complex plane of the complex number $a+ib=Ae^{i\phi}$ then we can draw

with

$$a=A\cos{\phi}$$

$$b=A\sin{\phi}$$

Then

$$f(t)=A(\cos{\phi}\cos{\omega t}+\sin{\phi}\sin{\omega t})$$

$$=A\cos{(\omega t-\phi)}$$

All we did was change our two unknown coefficients $a$ and $b$ to two other unknown coefficients $A$ and $\phi$.

In fact, in terms of the differential equation that has $f(t)$ as solution, we can either specify $f(0)$ and $\dot{f}(0)$ as initial conditions or, since these are functions of $a$ and $b$, respectively, and these are functions of $A$ and $\phi$, we can specify initial conditions in terms of the latter two parameters.

Next, let's see if we can get a more physical interpretation of what is happening.

Let's see why $a$ and $b$ can be thought of as polar coordinates in a physical situation.

$e^{-i\omega t}=\cos{\omega t}-i\sin{\omega t}$ represents clockwise uniform circular motion in the complex plane.

Let $a+ib=Ae^{i\phi}\in\mathbb{C}$.

$z(t)=(a+ib)e^{-i\omega t}=Ae^{-i(\omega t-\phi)}$ changes the amplitude and adds a phase to the motion.

The real part (ie, the "x" coordinate in the complex plane) is (I am omitting the calculations to obtain the real part, but it is just basic complex algebra):

$$a\cos{\omega t}+b\sin{\omega t}$$

which we know to be simple harmonic motion with initial position $a$ and initial velocity $b\omega$.

But note that

$$\text{Re}(z(t))=\text{Re}(Ae^{-i(\omega t-\phi)})$$

$$=A\cos{(-\omega t+\phi)}$$

$$=A\cos{(\omega t-\phi)}$$

Thus,

$$a\cos{\omega t}+b\sin{\omega t}=A\cos{(\omega t-\phi)}$$

What we have shown is that clockwise uniform circular motion in the complex plane with amplitude and phase given by $a+ib$ (which also specifies the initial position of the motion in the complex plane) has real coordinate that undergoes SHM on the real axis. This SHM is given by two equivalent expressions that we can equate, as we see in the last equation above.

So, in general, when we see an expression of the form $a\cos{\omega t}+b\sin{\omega t}$, we know that it can be thought of as not only SHM, but also the $x$-coordinate of a clockwise uniform circular motion in the complex plane with initial position given by $a+ib$.

Thus, when converting this expression to the amplitude-phase form, we can draw a triangle representing the complex number $a+ib$ with argument $\phi$ and amplitude $A$, and we can immediately interpret the final form as an amplitude $A$ times a cosine shifted by a phase, where the cosine by itself represents the $x$ coordinate of a uniform circular motion of radius 1 with no phase angle.