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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 $a=\frac{-eE}{m}$

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

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  • $\begingroup$ You have an initial condition such that the velocity and acceleration at the initial time is zero, so how can the particle ever move? $\endgroup$ – Farcher Nov 17 '19 at 19:10
  • $\begingroup$ Only velocity is 0 at the initial time. $\endgroup$ – Anon Nov 17 '19 at 19:17
  • $\begingroup$ @Farcher: the particle can move, as the third temporal derivative of coordinate does not vanish. This can be a driven oscillator. $\endgroup$ – akhmeteli Nov 17 '19 at 19:21
  • $\begingroup$ $a = -\omega^2 x$ 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. $\endgroup$ – ja72 Nov 17 '19 at 20:58
  • $\begingroup$ 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. $\endgroup$ – Aaron Stevens Nov 18 '19 at 13:55
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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 $a'(t) = a(t)$.

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  • $\begingroup$ But x has the form : $x=\frac{a_0}{\omega}t-\frac{a_0}{\omega^2}\sin(\omega t) $ $\endgroup$ – Anon Nov 18 '19 at 9:12
  • $\begingroup$ The the initial condition of $\dot{x}(0)=0$ won't be satisfied. $\endgroup$ – ja72 Nov 18 '19 at 13:48
  • $\begingroup$ But $v=\frac{a_0}{\omega}-\frac{a_0}{\omega}\cos(\omega t) $ so $v(0)= \frac{a_0}{\omega}-\frac{a_0}{\omega}=0$ $\endgroup$ – Anon Nov 19 '19 at 17:21
  • $\begingroup$ 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. $\endgroup$ – ja72 Nov 19 '19 at 17:53
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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?

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

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