Question #104535

Prove the solutions to this simple harmonic oscillation a=w^2x

Expert's answer

For this simple harmonic oscillation

"a=-\\omega^2x"

"a=\\frac{d^2x}{dt^2}"

"\\frac{d^2x}{dt^2}=-\\omega^2x"

Substitute "x(t)=e^{\\lambda t}"

We have

"e^{\\lambda t}(\\omega^2+\\lambda^2)=0"

"e^{\\lambda t}\\neq0"

"\\lambda=i\\omega" or "\\lambda=-i\\omega"

"x_1(t)=C_1e^{-i\\omega t}" and "x_2(t)=C_2e^{i\\omega t}"

The general solution is the sum of the above solutions

"x(t)=C_1e^{-i\\omega t}+C_2e^{i\\omega t}"

Using the Euler's identity we get

"x(t)=(C_1+C_2)\\cos(\\omega t)+i(-C_1+C_2)\\sin(\\omega t)"

In general

"x(t)=A_1\\cos(\\omega t)+A_2\\sin(\\omega t)"

or

"x(t)=A\\cos(\\omega t+\\phi)"

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