Answer to Question #104535 in Mechanics | Relativity for Ojugbele Daniel

Question #104535
Prove the solutions to this simple harmonic oscillation a=w^2x
1
Expert's answer
2020-03-05T09:36:34-0500

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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