# Simple harmonic motion position and velocity

It is given that the acceleration of a particle is $$a=a_0\sin(wt)$$ and $$v(0)=0.$$ Therefore, $$v=\frac{a_0}{\omega}-\frac{a_0}{\omega}\cos(\omega t)$$

$$x=\frac{a_0}{\omega}t-\frac{a_0}{\omega^2}\sin(\omega t)$$

But since it is SHM : $$a=-\omega^2 x$$

$$-\omega^2x=-a_0 \omega t+a_0\sin(\omega t)$$

And this is not equal to my given a . Is there any mistake?

Edit:

A particle is subject to an electric field $$E=E_0sin(wt)$$ which causes the particle an acceleration $$a=\frac{-eE}{m}$$

$$a_0=\frac{-eE_0}{m}$$

• You have an initial condition such that the velocity and acceleration at the initial time is zero, so how can the particle ever move? – Farcher Nov 17 '19 at 19:10
• Only velocity is 0 at the initial time. – Anon Nov 17 '19 at 19:17
• @Farcher: the particle can move, as the third temporal derivative of coordinate does not vanish. This can be a driven oscillator. – akhmeteli Nov 17 '19 at 19:21
• $a = -\omega^2 x$ only applies when $x$ has a zero value when $a$ is zero (initially). You need to shift $x$ a constant amount to get there in your case. – ja72 Nov 17 '19 at 20:58
• This is a good example of knowing where certain equations come from and understanding when they are true. Blindly applying equations will not get you anywhere in physics. – Aaron Stevens Nov 18 '19 at 13:55

The expression $$a = -\omega^2 x$$ is not true in general, but in every SHM problem there exists one inertial coordinate system which makes it true.

In your question with $$a = a_0 \sin(\omega t)$$ the general solution to the equations of motion are

$$\boxed{ x(t) = x_0 + \frac{a_0 + \omega\, v_0}{\omega} t - \frac{a_0}{\omega^2} \sin(\omega t) }$$ subject to the initial conditions $$x(0)=x_0$$ and $$v(0)=v_0$$.

The general solution does not obey $$\ddot{x} = -\omega^2 x$$

But consider a change in coordinate systems with $$x' = x -x_0 - \frac{a_0 + \omega\, v_0}{\omega} t$$ which makes the solution $$x'(t) = - \frac{a_0}{\omega^2} \sin(\omega t)$$ which does obey $$\boxed{ a = -\omega^2 x' }$$

For the specific case of $$x(0)=0$$ and $$v(0)=0$$ the necessary change in coordinate system is to a constant velocity coordinate of $$x' = \frac{a_0}{\omega} t$$ which does not violate Newton's requirement for inertial coordinate systems since the reference frame maintains constant velocity. In the new coordinate system the acceleration is unchanged $$a'(t) = a(t)$$.

• But x has the form : $x=\frac{a_0}{\omega}t-\frac{a_0}{\omega^2}\sin(\omega t)$ – Anon Nov 18 '19 at 9:12
• The the initial condition of $\dot{x}(0)=0$ won't be satisfied. – ja72 Nov 18 '19 at 13:48
• But $v=\frac{a_0}{\omega}-\frac{a_0}{\omega}\cos(\omega t)$ so $v(0)= \frac{a_0}{\omega}-\frac{a_0}{\omega}=0$ – Anon Nov 19 '19 at 17:21
• See edits in answer to clarify the issues. I might have confused the issue with my original post. My point is that a coordinate system with constant velocity is needed to make the SHM equation to be valid. – ja72 Nov 19 '19 at 17:53

One needs to see the entire wording of your problem. For all we know, this can be a driven oscillator, so the relation between the coordinate and acceleration does not have to hold.

EDIT (11/17/2019): So, according to your edit, the oscillator is indeed driven, then why do you expect $$a=-\omega^2 x$$ to hold?

• I updated my question. – Anon Nov 17 '19 at 19:29
• How do I know the oscillator is driven? – Anon Nov 17 '19 at 21:24
• @Anon : You said that it was driven by electric field. – akhmeteli Nov 17 '19 at 21:41
• So $−a_0ωt$ corresponds to $\frac{F(t)}{m}$ ? – Anon Nov 17 '19 at 22:29
• @Anon : Why is that? – akhmeteli Nov 17 '19 at 23:32