User:
Mathematical Pendulum
Storyboard 
In the case of a pendulum with a point mass, the potential energy is generated by raising the mass against the gravitational field as the pendulum deviates by a given angle.
ID:(1420, 0)
Oscillations with a mathematical pendulum
Description
A pendulum is described as a point mass $m$ hanging from a string that is attached to a pivot point and has a length $l$. It is called a mathematical pendulum because it is an abstraction of a physical pendulum, with the difference being that its mass is treated as a point mass.
ID:(7098, 0)
Mathematical pendulum
Description
A pendulum is defined by a point mass $m$ hanging from a string attached to a pivot of length $l$. It is referred to as a mathematical pendulum because it is an abstraction of a physical pendulum, where the mass is considered to be concentrated at a single point.
ID:(1180, 0)
Model
Top
Parameters
Variables
Calculations
to 
, then, select the variable: 
to 
Calculations
Calculations
Variable 
Given Calculate 
Target : Equation To be used 
Equations
$ E = K + V $
E = K + V
$ K =\displaystyle\frac{1}{2} m_i L ^2 \omega ^2$
K = m_i * L ^2* omega ^2/2
$ m_g = m_i $
m_g = m_i
$ \nu =\displaystyle\frac{1}{ T }$
nu =1/ T
$ \omega_0 = 2 \pi \nu $
omega = 2* pi * nu
$ \omega_0 = \displaystyle\frac{2 \pi }{ T }$
omega = 2* pi / T
$ \omega_0 ^2=\displaystyle\frac{ g }{ L }$
omega_0 ^2 = g / L
$ \omega = - \theta_0 \omega_0 \sin \omega_0 t $
v = - x_0 * omega_0 *sin( omega_0 * t )
$ V =\displaystyle\frac{1}{2} m_g g L \theta ^2$
V = m_g * g * L * theta ^2/2
$ E =\displaystyle\frac{1}{2} m_g g L \theta_0 ^2$
V = m_g * g * L * theta ^2/2
$ \theta = \theta_0 \cos \omega_0 t $
x = x_0 *cos( omega_0 * t )
ID:(15852, 0)
Total Energy
Equation
The total energy corresponds to the sum of the total kinetic energy and the potential energy:
| $ E = K + V $ |
ID:(3687, 0)
Kinetic energy of a mathematical pendulum
Equation
The kinetic energy of a rotating body is given by
| $ K_r =\displaystyle\frac{1}{2} I \omega ^2$ |
where $I$ is the moment of inertia and $\omega$ is the angular velocity. For a point mass $m$ rotating at a distance $L$ from an axis, the moment of inertia is
| $ I = m L ^2$ |
hence,
| $ K =\displaystyle\frac{1}{2} m_i L ^2 \omega ^2$ |
ID:(4515, 0)
Potential energy of a mathematical pendulum for small angles (1)
Equation
The gravitational potential energy of a pendulum is
| $ U = m g L (1-\cos \theta )$ |
which for small angles can be approximated as:
| $ V =\displaystyle\frac{1}{2} m_g g L \theta ^2$ |
The gravitational potential energy of a pendulum with mass
| $ U = m g L (1-\cos \theta )$ |
where
For small angles, the cosine function can be approximated using a Taylor series expansion up to the second term
$\cos\theta\sim 1-\displaystyle\frac{1}{2}\theta^2$
This approximation leads to the simplification of the potential energy to
| $ V =\displaystyle\frac{1}{2} m_g g L \theta ^2$ |
It's important to note that the angle must be expressed in radians.
ID:(4514, 1)
Potential energy of a mathematical pendulum for small angles (2)
Equation
The gravitational potential energy of a pendulum is
| $ U = m g L (1-\cos \theta )$ |
which for small angles can be approximated as:
| $ E =\displaystyle\frac{1}{2} m_g g L \theta_0 ^2$ |
| $ V =\displaystyle\frac{1}{2} m_g g L \theta ^2$ |
The gravitational potential energy of a pendulum with mass
| $ U = m g L (1-\cos \theta )$ |
where
For small angles, the cosine function can be approximated using a Taylor series expansion up to the second term
$\cos\theta\sim 1-\displaystyle\frac{1}{2}\theta^2$
This approximation leads to the simplification of the potential energy to
| $ V =\displaystyle\frac{1}{2} m_g g L \theta ^2$ |
It's important to note that the angle must be expressed in radians.
ID:(4514, 2)
Equality of inertial and gravitational mass
Equation
The masses that Newton used in his principles are related to the inertia of bodies, which leads to the concept of the inertial Mass ($m_i$).
Newton's law, which is linked to the force between bodies due to their masses, is related to gravity, hence known as the gravitational mass ($m_g$).
Empirically, it has been concluded that both masses are equivalent, and therefore, we define
| $ m_g = m_i $ |
Einstein was the one who questioned this equality and, from that doubt, understood why both 'appear' equal in his theory of gravity. In his argument, Einstein explained that masses deform space, and this deformation of space causes a change in the behavior of bodies. Thus, masses turn out to be equivalent. The revolutionary concept of space curvature implies that even light, which lacks mass, is affected by celestial bodies, contradicting Newton's theory of gravitation. This was experimentally demonstrated by studying the behavior of light during a solar eclipse. In this situation, light beams are deflected due to the presence of the sun, allowing stars behind it to be observed.
ID:(12552, 0)
Angular frequency of a mathematical pendulum
Equation
In the case of the mathematical pendulum
the energy can be expressed as
$E=\displaystyle\frac{1}{2}ml^2\omega^2+\displaystyle\frac{1}{2}mgl\theta^2$
and from this expression, we can obtain the angular frequency
| $ \omega_0 ^2=\displaystyle\frac{ g }{ L }$ |
The kinetic energy of the mathematical pendulum with mass $m$, string length $r$, and angular velocity $\omega$ is
| $ K =\displaystyle\frac{1}{2} m_i L ^2 \omega ^2$ |
and the gravitational potential energy is
| $ V =\displaystyle\frac{1}{2} m_g g L \theta ^2$ |
With $\theta$ representing the angle and $g$ the angular acceleration, the equation for the total energy is expressed as
$E=\frac{1}{2}m r^2 \omega^2 + \frac{1}{2}m g r \theta^2$
Given that the period is equal to
$T=2\pi\sqrt{\frac{m r^2}{m g r}}=2\pi\sqrt{\frac{r}{g}}$
we can relate the angular frequency as
| $ \omega_0 ^2=\displaystyle\frac{ g }{ L }$ |
ID:(4516, 0)
Angular frequency
Equation
The angular frequency ($\omega$) is with the period ($T$) equal to
| $ \omega_0 = \displaystyle\frac{2 \pi }{ T }$ |
| $ \omega = \displaystyle\frac{2 \pi }{ T }$ |
ID:(12335, 0)
Frequency
Equation
The frequency ($\nu$) corresponds to the number of times an oscillation occurs within one second. The period ($T$) represents the time it takes for one oscillation to occur. Therefore, the number of oscillations per second is:
| $ \nu =\displaystyle\frac{1}{ T }$ |
Frequency is indicated in Hertz (Hz).
ID:(4427, 0)
Relación frecuencia angular - frecuencia
Equation
Como la frecuencia angular es con angular frequency $rad/s$, period $s$ and pi $rad$ igual a
| $ \omega_0 = \displaystyle\frac{2 \pi }{ T }$ |
y la frecuencia con frequency $Hz$ and period $s$ igual a
| $ \nu =\displaystyle\frac{1}{ T }$ |
se tiene que con frequency $Hz$ and period $s$ igual a
| $ \omega_0 = 2 \pi \nu $ |
| $ \omega = 2 \pi \nu $ |
ID:(12338, 0)
Oscillation amplitude
Equation
With the description of the oscillation using
| $ z = x_0 \cos \omega_0 t + i x_0 \sin \omega_0 t $ |
the real part corresponds to the temporal evolution of the amplitude
| $ \theta = \theta_0 \cos \omega_0 t $ |
| $ x = x_0 \cos \omega_0 t $ |
ID:(14074, 0)
Swing speed
Equation
When we extract the real part of the derivative of the complex number representing the oscillation
| $ \dot{z} = i \omega_0 z $ |
whose real part corresponds to the velocity
| $ \omega = - \theta_0 \omega_0 \sin \omega_0 t $ |
| $ v = - x_0 \omega_0 \sin \omega_0 t $ |
Using the complex number
| $ z = x_0 \cos \omega_0 t + i x_0 \sin \omega_0 t $ |
introduced in
| $ \dot{z} = i \omega_0 z $ |
we obtain
$\dot{z} = i\omega_0 z = i \omega_0 x_0 \cos \omega_0 t - \omega_0 x_0 \sin \omega_0 t$
thus, the velocity is obtained as the real part
| $ v = - x_0 \omega_0 \sin \omega_0 t $ |
ID:(14076, 0)