Question

The amplitude of the oscillation

Answer 1

I have given this equation:
\[\frac{1}{2}m \dot x^2+\frac{1}{2}k(x-a)^2=E\] where m is mass, k iz positive constant, (x-a) is extension, E is constant (this equation for total mechanical energy of a given motion).

Firstly, how can I conclude that this is periodic motion? Is that because both terms at left side are positive and E is constant, so if value of one term increases, the value of another term has to decrease?

What is the amplitude of such oscillation?

Answer 2

whats a/

Answer 3

A particle is attached at one end of extensible spring of natural length a and stiffness coefficient k. So, a is natural length of a spring. It moves on smooth horizontal surface. Potential energy is \[\frac{1}{2}k(x-a)^2\]
kinetic energy is\[\frac{1}{2}m \dot x^2\]

Answer 4

the first is a Kinetic Energy while the second is a Potential Energy. It's natural that the system oscillates about the point 'a' which is mean position where PE is minimum and and KE is maximum.

To find the equation of motion, use Lagrangian mechanics
\[ L = T - V = \frac{1}{2} m \dot x ^2 - \frac 1 2 k (x-a)^2 \]

let x' = x - a, you will note that \( \dot x' = \dot x\)
the equation of motion will be
\[ \frac{d}{dt} \left( \frac{\partial L }{\partial \dot x' }\right) - \frac{\partial L }{\partial x' } = 0\]

Answer 5

this should reduce to
\[ m \ddot x' + k x' = 0\]
which is equation of harmonic oscillator, upon solving this you should get
\[x' = x-a = A \cos ( \omega t + \phi) \]
the value of A and phi depends on your initial condition.

Answer 6

How can I do this without Lagrangian because we learned Energy before we knew anything about Lagrange's mechanics? So I shouldn't use it in this problem.

Answer 7

use newton's second law
F = dV/dx = dp/dt
KE = p^2/2m

Answer 8

woops!!
\[ F = - \frac{dV}{d(x-a)} = - k (x-a) = \frac{d}{dt } \sqrt{\left( 2m * \frac 1 2 m \dot x^2 \right)} = m \ddot x \]
which you get
\[ m \ddot x + k (x-a) = 0\]
let x' = x-a again and same procedure

Answer 9

what is \[\sqrt{(2m*\frac{1}{2}m \dot x^2)}\]?

Answer 10

you know
\[ KE = \frac{p^2}{2m} \implies p = \sqrt{2m \times KE }\]

Answer 11

what is p? impuls?

Answer 12

what is V/
?

Answer 13

momentum ... the rate of change of momentum gives force. V is potential energy

Answer 14

* the rate of change of momentum gives force as well the negative gradient of potential gives force. just equate them and solve that DE.

Answer 15

it's ok now, thank you :)

Answer 16

very good

Answer 17

thnks

Answer 18

yw