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In the case of the spring the force is proportional to the elongation of the spring so that the equations of motion are linear and the frequency of the oscillation is independent of the amplitude. This is the key to generate an oscillation that does not depend on the fact that the friction decreases over time. This is why old clocks used (circular) springs to generate stable oscillations to measure elapsed time.

>Model

ID:(1425, 0)

Equation

>Top, >Model

The kinetic energy of a mass $m$

can be expressed in terms of momentum as

$ K =\displaystyle\frac{ p ^2}{2 m_i }$

Conditions

Variables

$m_i$

Inertial Mass

$kg$

$p$

Momento

$kg m/s$

$K$

Space Time Position

$J$

Relation

Development

Since kinetic energy is equal to

$ K_t =\displaystyle\frac{1}{2} m_i v ^2$

and momentum is

$ p = m_i v $

we can express it as

$K_t=\displaystyle\frac{1}{2} m_i v^2=\displaystyle\frac{1}{2} m_i \left(\displaystyle\frac{p}{m_i}\right)^2=\displaystyle\frac{p^2}{2m_i}$

or

$ K =\displaystyle\frac{ p ^2}{2 m_i }$

ID:(4425, 0)

Equation

>Top, >Model

En el caso elástico (resorte) la fuerza es



la energía

se puede mostrar que en este caso es

$ V =\displaystyle\frac{1}{2} k x ^2$

Conditions

Variables

$x$

Elongation of the Spring

$m$

$k$

Hooke Constant

$N/m$

$V$

Potential Energy

$J$

Relation

Development

En el caso elástico (resorte) la fuerza es



con k la constante del resorte y x la elongación/compresión del resorte. La variación de la energía potencial es

$ \Delta W = \vec{F} \cdot \Delta\vec{s} $

\\n\\nLa diferencia\\n\\n

$\Delta x = x_2 - x_1$

\\n\\ncorresponde al camino recorrido por lo que\\n\\n

$\Delta W=k,x,\Delta x=k(x_2-x_1)\displaystyle\frac{(x_1+x_2)}{2}=\displaystyle\frac{k}{2}(x_2^2-x_1^2)$

y con ello la energía potencial elástica es

$ V =\displaystyle\frac{1}{2} k x ^2$

ID:(3246, 0)

Equation

>Top, >Model

For the case of a mass oscillating with a spring, the energy as a function of momentum $p$ and position $q$ is

$ E_s =\displaystyle\frac{ p ^2}{2 m_i }+\displaystyle\frac{ k }{2} q ^2$

Conditions

Variables

$E_k$

Energy of a spring system

$J$

$k$

Hooke Constant

$N/m$

$m_i$

Inertial Mass

$kg$

$p$

Momento

$kg m/s$

$s$

Posición (vector)

$m$

Relation

Development

The kinetic energy as a function of momentum is given by

$ K =\displaystyle\frac{ p ^2}{2 m_i }$

and the potential energy as a function of height is

$ V =\displaystyle\frac{1}{2} k x ^2$

So, if we express the elongation as the position

$x = q$

we obtain

$ E_s =\displaystyle\frac{ p ^2}{2 m_i }+\displaystyle\frac{ k }{2} q ^2$

The equation can be expressed in an adimensional form as

$1=y^2 + x^2$

where

$x=\displaystyle\frac{q}{\sqrt{2E/k}}$

, and

$y=\displaystyle\frac{p}{\sqrt{2m_iE}}$

solving for y, we obtain

$y=\pm\sqrt{1-x^2}$

Its representation in the xy plane is shown below

ID:(1187, 0)

Equation

>Top, >Model

Como la oscilación cumple las leyes físicas se puede hacer uso del hecho que el area debajo de la curva velocidad vs tiempo el camino recorrido para determinar el perido. Como la velocidad es\\n\\n

$\displaystyle\int_0^{T/2}v(t)dt=\sqrt{\displaystyle\frac{2E}{m}}\displaystyle\int_0^{T/2}\cos \displaystyle\frac{2\pi t}{T}dt=\sqrt{\displaystyle\frac{2E}{m}}\displaystyle\frac{T}{\pi}$

\\n\\ny el camino entre un mínimo a un máximo de una elongación, lo que ocurre entre el tiempo 0 y T/2 es igual a\\n\\n

$x_{max}-x_{min}=2\sqrt{\displaystyle\frac{2E}{k}}$

se tiene que

$ T =2 \pi \sqrt{\displaystyle\frac{ m_i }{ k }}$

Conditions

Variables

$k$

Hooke Constant

$N/m$

$m_i$

Inertial Mass

$kg$

$T$

Period

$s$

$\pi$

Pi

3.1415927

$rad$

Relation

ID:(7106, 0)

Equation

>Top, >Model

One of the systems it depicts is that of a spring. This is associated with the elastic deformation of the material from which the spring is made. By "elastic," we mean a deformation that, upon removing the applied stress, allows the system to fully regain its original shape. It's understood that it doesn't undergo plastic deformation.

Since the energy of the spring is given by

$E=\displaystyle\frac{1}{2}m_i v^2+\displaystyle\frac{1}{2}k x^2$

the period will be equal to

$T=2\pi\sqrt{\displaystyle\frac{m_i}{k}}$

and thus, the angular frequency is

$ \omega_0 ^2=\displaystyle\frac{ k }{ m_i }$

Conditions

Variables

$\omega_0$

Frecuencia angular del resorte

$rad/s$

$k$

Hooke Constant

$N/m$

$m_i$

Inertial Mass

$kg$

Relation

Development

Since the kinetic energy depends on the mass $m$ and velocity $v$, it is given by

$ K_t =\displaystyle\frac{1}{2} m_i v ^2$

and the potential energy of the spring, which depends on the spring constant $k$ and elongation $x$, is

$ V =\displaystyle\frac{1}{2} k x ^2$

Thus, the total energy is expressed as

$E=\displaystyle\frac{1}{2}m_i v^2+\displaystyle\frac{1}{2}k x^2$

As the period is

$T=2\pi\sqrt{\displaystyle\frac{m_i}{k}}$

we can calculate the angular frequency as

$\omega_0=\displaystyle\frac{2\pi}{T}=\sqrt{\displaystyle\frac{k}{m_i}}$

which implies

$ \omega_0 ^2=\displaystyle\frac{ k }{ m_i }$

ID:(1242, 0)

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Video

Video: Oscillators on a Spring