71
As humans we oscillate left and right when we walk because we have two legs. You can get a resonance when the length of the cord is such that your pace matches the period of the swing.
(Like pushing a child on a swing a little higher each time they approach you.)
Whilst walking we also oscillate up and down - this can also contribute to driving the resonance....
66
The answer to this question has significant overlap with my answer on piano tuning. There, I discuss how a thick wire has an extra restoring force, in addition to its tension, from its resistance to bending. This modifies the usual wave equation to
$$v^2 \frac{\partial^2 y}{\partial x^2} - A \frac{\partial^4 y}{\partial x^4} = \frac{\partial^2 y}{\partial t^...
39
The balloon has a very small mass and friction is large (large surface area), so the oscillation is very damped.
32
A simple pendulum does not strictly show simple harmonic motion unless you allow some approximations and uncertainties. It approximately behaves as a harmonic oscillator for small amplitudes.
An object is said to be executing simple harmonic motion (no damping; not a forced oscillation) if and only if it satisfies the following condition:
$$\frac{d^2 \phi}{...
27
Chaotic is not the same as random. A chaotic system is entirely deterministic, while a random system is entirely non-deterministic. Chaotic means that infinitesimally close initial conditions lead to arbitrarily large divergences as the system evolves. But it's impossible, practically speaking, to reproduce the same initial conditions twice. Given ...
25
In the centre of a bowl there is equilibrium.
Put a ping pong ball in it. If this ball is ever shaken slightly away from equilibrium, it will immediately roll back. A "shake-proof" equilibrium is called stable.
Now, turn the bowl around and put the ball on the top. This is an equilibrium. But at the slightest shake, the ball rolls down. This "non-shake-...
23
When you pluck the string you excite many many overtones, not just the fundamental. You can observe this by suppressing the fundamental. Pluck the string while holding a finger lightly at the center of the string. That point is an antinode for the fundamental and all odd harmonics, but a node for the even harmonics. Putting your finger at that point ...
22
Tied balloons do behave like a pendulum, you only need really massive ones:
You can see it live in a video.
Hot air balloons have such a big amount of..erm...hot air that during the start you can
expect oscillations because while the surface area is big, the mass inside is so big that the dampening is low enough. They are not exactly like a pendulum ...
20
As knzhou identifies, the key difference between vibrations of a free beam and a string is that the restoring force is now provided by bending moments (proportional to $\frac{d^4y}{dx^4}$) rather than linear tension (proportional to $\frac{d^2y}{dx^2}$).
The consequence, as this source shows, is that for free beams like the bars of the glockenspiel, angular ...
19
There seem to be a lot of human body mechanical models, such as this one:
As for applications, I have heard that sub-audio frequency vibrations have been considered as nonlethal weapons for riot control.
19
This is a subtle issue! Your intuition is correct (a driving at $f_0/2$ should be very effective) even though the graph seems to contradict this. The reason is that the graph displays the response to a sinusoidal driving force. If you indeed drove the mass sinusoidally at frequency $f_0/2$, it would indeed be ineffective -- you'd be holding onto the mass and ...
18
I think Crawford's "Waves" is an incredible book - full of insight and clearly written by someone who loves the material. I used it for my waves-course sophomore year, and I think it's too bad it's out of print now.
If you want something more theoretical, though, try Howard Georgi's book.
Also, I'll second A.P.French's "Vibrations and Waves".
Also, David ...
16
One of the big reasons not discussed above is Fourier theory -- any function $f(x)$ can be expressed in the form $f(x) = \int dk\, A(k)e^{ikx}$, which basically means that any function can be decomposed into an infinite sum of sines and cosines. Since this is the case and dealing with sine and cosine is mathematically simpler than the general case of ...
16
Usually, the (sinusoidal) driven harmonic oscillator is damped, and the first two parts of your solution (which depend on the initial conditions, while the third term does not) are transient, i.e. not relevant after a short time. That the solution
$$ x(t) = \frac{F_0\sin(\omega t)}{m(\omega_0^2 - \omega^2)}$$
cannot be the "full" solution to the equation of ...
16
In a very mathematical sense, more often than not a mode refers to an eigenvector of a linear equation.
Consider the coupled springs problem
$$\frac{d}{dt^2} \left[ \begin{array}{cc} x_1 \\ x_2 \end{array} \right]
=\left[ \begin{array}{cc}
- 2 \omega_0^2 & \omega_0^2 \\
\omega_0^2 & - \omega_0^2
\end{array} \right]
\left[ \begin{array}{cc} x_1 \\ x_2 ...
12
Because cycles and oscillations and things with periodicity, are all intimately related to the circle. And $sin$ and $cos$ are defined based on the circle.
12
That the true period of a pendulum is longer than the period found using the harmonic oscillator (HO) approximation can actually be deduced just from the fact that $\sin\theta<\theta$ for $\theta>0$. That means that the restoring torque on the pendulum bob is always less than in the approximate HO system. Weakening the restoring force automatically ...
11
These kinds of proportionality questions are often best answered with dimensional analysis. You want to know a form a quantity with the units of time in terms of what you have.
You have a quantity $k$ with units $\frac{\text{Energy}}{\text{Distance}^3} = \frac{\text{Mass}}{\text{Distance} \times \text{Time}^2}$. You also have the mass $m$ (units of Mass) ...
11
Suppose we take two identical pendulums. The green one is undamped and the red one is damped:
The force $F_g$ on the green pendulum bob is (approximately) given by the usual simple harmonic oscillator law:
$$ F_g = -kx $$
where $x$ is the displacement, and the acceleration is just $F_g/m$.
Now consider the red pendulum bob. This is damped, so the force ...
11
First I'll try to explain why the amplitude vs frequency diagram only has one maximum, then I'll go back to why this seems to contradict your intuition.
Let's take the simplest forced oscillator formula, with no damping (this won't affect our conclusion), for instance that of a spring undergoing a force $F$ :
\begin{equation}
x''(t) + \omega_0^2 x(t) = F(t)...
10
It would depend on damping effects being taken into account or not.
Invoking Newton's 2nd Law of motion, a differential equation for the motion of a damped harmonic oscillator can be written (including an external, sinusoidal driving force term):
$m\frac{d^2x}{dt^2}+2m\xi\omega_0\frac{dx}{dt}+m\omega_0^2x=F_0\sin\left(\omega t\right)$
Where $m$ is the ...
10
The oscillator frequency $\omega$ says nothing about the actual oscillator phase. Let us suppose that your oscillator oscillates freely like this: $$x(t) = A_0\cdot\cos(\omega t + \phi_0),\; t<0.$$ At $t=0$ it has a phase $\phi_0$. Depending on its value the oscillator can be moving forward or backward with some velocity. If you switch your external force ...
10
Sometimes, a good figure is worth more than a thousand equations :)
I numerically integrated the following equation of motion for a physical pendulum:
$$
I\ddot{\theta} + mgL\sin(\theta) + \frac12\mathrm{sgn}({\dot{\theta}})L\rho_{\mathrm{air}}C_DS(L\dot{\theta})^2 + \zeta\dot{\theta} + \gamma\theta = 0
$$
with $\mathrm{sgn()}$ the signum function.
The ...
10
A literal response to your title question would simply be "because in the physical world, oscillations behave in ways consistent with $\sin$ and $\cos$." Of course, one then wonders why these functions are so ubiquitous.
Depending on your level of physics background, you may be familiar with the harmonic oscillator - that is, a system for which there exists ...
10
The effect of gravity is only to shift the equilibrium point, so at equilibrium (at rest), a vertical spring will be extended as compared with the same spring in a horizontal position. But this does not affect the period.
The equation for the dynamics of the spring is $m\frac{d^2x}{dt^2}=-kx+mg$. You can change the variable $x$ to $x'=x+mg/k$ and get $m\...
10
Mathematical demonstration
It's straightforward to see why this happens if you use a bit of linear response theory.
Consider a generic damped harmonic oscillator.
There are three forces, the restoring force $F_\text{restoring} = - k x(t)$, the friction force $F_\text{friction} = - \mu \dot{x}(t)$, and the driving force $F_\text{drive}(t)$.
Newton's law says ...
9
The answers currently posted are ignoring a few important details so I'm going to give my own.
I may rehash some things already said.
To make everything absolutely clear I write here a complete derivation of the forced damped oscillator with emphasis on the role of the $Q$ factor.
Basic equations
Consider the equation of motion of a forced, damped harmonic ...
9
This is an answer by an experimentalist who had been fitting data with mathematical models since 1968.
When fitting data one goes to the simplest mathematical models. When the data display variations in time and space the Fourier expansion is extremely useful because it gives the frequencies and amplitudes that will fit a periodic data set. One gets as ...
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