SHM phase relations between v,a,y

Question

I was reading about the phase relation between the displacement , velocity and acceleration of a particle executing SHM.I understood that all three differ in phase by π/2 radian respectively . In my textbook it is written in addition to the above that
the velocity in SHM is in phase with acceleration , when particle is moving from extreme position to mean position and it is in opposite phase with acceleration when particle is moving from mean position to extreme position
How?

Rohit · Accepted Answer

The differential equation for SHM in 1D is given by
begin{equation}
ddot{y}+omega^2y=0
end{equation}
Its solution can be written as
begin{equation}
y(t)=Asin(omega t+phi)
end{equation}
where $A$ and $phi$ is determined by the initial conditions (say $phi=0$, which means that oscillator starts from equilibrium position). Then,
begin{equation}
y(t)=Asin(omega t)
end{equation}
begin{equation}
v(t)=dot{y}(t)=omega Acos(omega t)
end{equation}
begin{equation}
a(t)=dot{v}(t)=-omega^2 A sin(omega t)
end{equation}
So the difference between $v$ and $a$ is always $pi/2$ (rather $3pi/2$). See the plot for $omega=2pi$ (black curve is $v$ and red curve is $a$), in going from equilibrium position to extreme, both $v$ and $a$ have opposite signs (blue region) and going from extreme to mean position they have the same sign (green region). Probably this is what you mean when you say 'phase'.

SHM phase relations between v,a,y

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