Question #102594

Show that x=Asinwt+Acoswt, satisfies the differential equation md^2x/dt^2 + kx=0

Expert's answer

"x=A\\sin \\omega t+A\\cos \\omega t"

"\\frac{dx}{dt}=A\\omega\\cos\\omega t-A\\omega\\sin\\omega t"

"\\frac{d^2x}{dt^2} =\\frac{d}{dt}(A\\omega\\cos\\omega t-A\\omega\\sin\\omega t)= -A\\omega^2 \\sin\\omega t-A\\omega^2\\cos\\omega t"

Substitute x into this equation: "m\\frac{d^2x}{dt^2}+kx=0"

"m( -A\\omega^2 \\sin\\omega t-A\\omega^2\\cos\\omega t)+k( A\\sin\\omega t+A\\cos\\omega t)=0"

"A(k-m\\omega ^2)(\\sin\\omega t+\\cos \\omega t)=0" for all t

Therefore, x satisfies the differential equation "m\\frac{d^2x}{dt^2}+kx=0"

if "A=0" or "k-m\\omega ^2=0."

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