Answer to Question #102594 in Mechanics | Relativity for Ojugbele Daniel

Question

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

Answer 1

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

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

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

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 $k-m\omega ^2=0.$