51
The human gait has a natural bobbing motion, with the head moving slightly up-and-down and side-to-side. The side-to-side motion (swinging on an axis parallel to the nose) turns the ponytail into a natural pendulum which swings back and forth, since this plane of motion is gravitationally symmetric and has nothing to stop the swing. Small driving forces can ...
11
I think the longitudinal oscillations of a ponytail are quickly damped (more precisely overdamped), since they involve layers of hair sliding along each other, as well as inelastic collisions of the back of the neck. On the other hand, the transversal oscillations require merely twisting the ponytail near the elastic band holding it together.
I think it ...
5
"Vibration" just refers to oscillatory motion around an equilibrium point. $\sin^2(x)$ certainly satisfies this definition.
Also it's worth pointing out that $\sin^2(x)$ actually is a sinusoidal function
\begin{equation}
\sin^2(x) = \frac{1 - \cos(2x)}{2}
\end{equation}
5
Not a complete answer, but a few notes.
There is a significant amount of rotational force going on in the head and neck during running, so much that humans (as well as other running animals) have specific adaptations to help manage them. Cf "that pig can't hold its head still"[1].
I can at least find one source[2] finding that the head has a yaw (...
3
Please note that mass $m$ and the spring constant $k$ are both the same on earth as they are on the moon. Consequently, the time period of the spring does not depend upon differences due to the acceleration due to gravity. Hence it will not change when it’s taken to the moon.
I think the value of 𝑘 for a spring hung upside down depends on gravitational ...
2
To first order, a pendulum simple harmonic motion and it's pendulum does not depend on the amplitude of its swing. However, the motion of a pendulum is not exactly simple harmonics motion.
To calculate the time period, you need to solve what is called elliptic integral. However you can approximate it, In the following case if you work out it is turn out ...
2
As @nuclear-hoagie said, the ponytail is basically a pendulum, so vibrations up and down are not really possible (or are much more complicated) as the hair would have to elongate itself in order to store the kinetic energy from the hair falling down so it cannot release it later. However the sideways and back and forth vibrations can be maintained almost for ...
2
Your considerations are correct. The particle is oscillating if the force is negative for $x > 1$ and positive for $x < 1$. However, this is only true for option 3 as can be seen if you plot the force $F(x)$ against $x$.
1
Assuming you are considering the forces that act along the pendulum bob's path, then the following is true for small angles:
Force vs. angular displacement
The graph is a straight line: $$F=-mg \theta\tag{1}.$$
Force vs time
Recall the simple pendulum equation $$\theta(t)=\theta_0\cos(\omega t+\phi)\tag{2}$$ where $\omega=\sqrt{g/L}$. Now substitute (2) into ...
1
Start with conservation of energy
$$
E=\frac{m}{2}\ell^2\dot{\theta}^2+mg\ell(1-\cos\theta)\, . \tag{1}
$$
The potential energy $mg\ell(1-\cos\theta)$ is not exactly quadratic in $\theta^2$, with the result that the period will generally depend on the amplitude of the motion. This is a pretty generic feature of all potentials, and it’s only for the ...
1
The differential equation of the simple pendulum gets a little messy, since it cannot be solved in closed form using elementary functions. The standard
analytical solution involves a complete elliptic integral of the first kind.
However, as that Wikipedia article mentions, there's a lovely iterative algorithm orininally due to Gauss which uses the arithmetic-...
1
If you want to know just for a small time you can approximate $sin(\theta)$ using Taylor series-
$$\Rightarrow sin(\theta)\approx sin(\theta_0)+(\theta-\theta_0)cos(\theta_0) $$
$$\frac{d^2x}{dt^2}+\frac{gcos(\theta_0)x}{l}+\frac{gsin(\theta_0)}{l}=0 $$
where $x=\theta-\theta_0$. Now if you assume $x$ as a power series in $t$, $x=\Sigma_{i=0}^{\infty}a_ix^i$ ...
1
You can describe the position of the pendulum with one degree of freedom, the angle $\theta$ which the pendulum mass makes with the vertical. Since $\theta$ is an angle, we identify $\theta$ and $\theta+2\pi$, meaning that (for example) $\theta=0$ and $\theta=2\pi$ are really the same physical point. Let's choose a convention where $\theta=0$ is the ...
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