# Mechanical oscillations question ? plz help!?

Given a horizontal spring of stifness K and to which is attached an object of mass m,

X" + k/M X = 0 2nd order differential equation that characterises the motion of the object

How can you prove that X = x sin (omega .t + phi ) is a solution of the above diff equation ? knowing that X is the elongation, x is the maximum elongation , omega is the angular frequency = sqrt (k/M) , t is the time, and phi is the initial angle

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• Anonymous

That depends what you know already.

If you know how to solve linear ODEs with constant coefficients, the general solution is "obviously"

X = A sin(omega t) + B cos(omega t)

where A and B are arbitrary constants.

You can rearrange that that into

X = x sin(omega t + phi)

where x = sqrt (A^2 + B^2) and tan phi = B/A

If you haven't studied ODEs, you can verify the solution is true by differating to find X" and substituting in the equation.

Or you can solve it by multiplying the equation by 2X'

Writing k/M = w^2 to make the typing easier,

X" + w^2 X = 0

2X"X' + (w^2) 2XX' = 0

d/dt(X'^2) + w^2 d/dt(X^2) = 0

Integrating,

X'^2 + w^2 X^2 = C where C is an arbitrary constant

The LHS is the sum of two squares so C > 0.

It's easier to write C = w^2 A^2 where A is another arbitrary constant.

X'^2 = w^2(A^2 - X^2)

X' = +/- w sqrt(A^2 - X^2)

That has solutions X = A sin wt or X = A cos wt depending on whether you take the + or - sign, so the general solution is X = A sin wt + B cos wt, as before.

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• take the derivative twice and substitute into the original equation. it's a matter of trial and error.

so X=x sin(omega t + phi),

so X'=- x cos (omega t + phi) (omega)

so X''=-x sin (omega t +phi)(omega)^2

subsitute into the orignal equation, -x sin (omega t + phi)(omega)^2+k/m(x sin(omega t + phi)=0

or (-(omega^2)+k/M)(x sin(omega t +phi))=0

so -omega^2+k/M=0, or omega=sqrt(k/M).....qed

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