# Driven harmonic oscillator: Why does the phase of the driver have such a big impact on the solution?

Consider the equation for a driven harmonic oscillator,
$$frac{d^2 x}{d t^2} + omega_0^2x(t) = f(t),$$
with initial conditions
$$x(0)=left.frac{dx}{dt}right|_{t=0}=0.$$
If we assume $$omega_0neomega$$ then
$$x(t)=x_0omega^2begin{cases} frac{omega_0sin(omega t) – omega sin(omega_0 t)}{omega_0(omega_0^2 – omega^2)}, & f(t)=x_0omega^2sin(omega t), \ frac{cos(omega t) – cos(omega_0 t)}{omega_0^2 – omega^2}, & f(t)=x_0omega^2cos(omega t). end{cases}$$
In the limit $$omegarightarrow infty$$,
$$x(t)rightarrow x_0begin{cases} frac{omega}{omega_0}sin(omega_0 t), & f(t)=x_0omega^2sin(omega t), \ cos(omega_0 t) – cos(omega t), & f(t)=x_0omega^2cos(omega t). end{cases}$$
This shows that depending on the phase of the driver the amplitude can be infinite or finite. Do you know a physical reason why this might be, or have I made a mistake in the maths somewhere?

Mathematics Asked on November 12, 2021

Consider the easier differential equation $$frac{d^2x}{dt^2} = f(t)$$ with $$x(0) = frac{dx}{dt}Big|_{t=0} = 0$$. If $$f(t) = x_0 omega^2 sin(omega t)$$ we get $$x(t) = x_0 omega t - x_0 sin(omega t) .$$ If $$f(t) = x_0 omega^2 cos(omega t)$$ we get $$x(t) = x_0 - x_0 cos(omega t) .$$ Again, the phase of the driving term determines whether the amplitude blows up or stays bounded.

It basically boils down to the constant terms. In the first case, there is a net velocity in the motion, which causes the solution to drift to infinity. In the second case, you 'get lucky' and the drift is exactly zero.

Answered by Stephen Montgomery-Smith on November 12, 2021

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