$$F(x)= -kx = ma(x) $$ Elastic force equation

I solve this differential equation to find the equation of motion:

$$ -kx =m\frac{d^2x}{dt^2} $$

$$ x'' +\frac{k}{m}x=0 $$

$λ^2 = -\frac{k}{m} $

$ λ = \pm i\sqrt{\frac{k}{m}}$

$ x(t) = c_1\cos(t\sqrt{\frac{k}{m}}) + c_2\sin(t\sqrt{\frac{k}{m}}) $ equation of motion

$ k $ is $\rm[M][T]^{-2} $ so I can write $ ω =\sqrt{\frac{k}{m}}$ and call it angular frequency $[T]^{-1} $

$ x(t) = c_1\cos(tω) + c_2\sin(tω) $

$c_1$ and $c_2$ are $[\rm L]^{1}$, but what they represent?, the equation should depend only from $x_0$ or I am wrong?

Then I can assume:

$c_1 = A\sin(φ)$

$c_2 = A\cos(φ)$

So using trigonometric addition formulas I can simplify the motion equation to:

$x(t) = A\sin(tω+φ)$ That is a Simple harmonic motion

A is $[\rm L]^1$, and I suppose $ A = x_0 $, it's correct, or I am wrong?

$ φ $ should be the phase, but I am not able to understand how it's related to force equation, how can I determine $φ$? What is physical meaning of $φ$?

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    $\begingroup$ $\phi$ represents the initial conditions of the system. It unequivocally tells us about what initial position the oscillator had and what was its velocity then. But I think the title of your question is quite misleading if the content of your question is what you want to ask. $\endgroup$ – Dvij D.C. Jun 19 '17 at 10:10
  • $\begingroup$ Sorry for bad title, but I am unable to find a better, if you have a better suggestion. I've nothing against put this question on hold, since it's already well answered. But I don't approve the reason: 1) This is not an homework 2) I put lot of effort solving the differential equation by myself, trying to understand the final formula. $\endgroup$ – Stefano Balzarotti Jun 20 '17 at 13:24
  • $\begingroup$ *for title, if you have a suggestion, I'll change. $\endgroup$ – Stefano Balzarotti Jun 20 '17 at 13:31
  • $\begingroup$ In my physics book, there are the two formula (Hooke's law and motion equation) without any step to correlate them. I've done by myself all steps to understand the formula. I've done dimensional analysis to understand the meaning of $A$, but i was unable to understand the meaning of $φ $, and I had need a confirm if steps and my final conclusion are correct. So you can't tell me that I haven't show some effort. $\endgroup$ – Stefano Balzarotti Jun 20 '17 at 13:41

As you stated correctly, the two equations $$x(t)=c_1 \cos(t\omega)+c_2 \sin(t\omega)$$ and $$x(t)=Asin(t\omega + \phi)$$ are equivalent and both are the general solution for the problem. In order to get from the general solution to a particular solution, you need to impose boundary conditions,e.g. \begin{align} x_0 = x(t=0s) = 1 cm\\ v_0 = v(t=0s) = 0 m/s \end{align} Both general solution will yield the same answer if they are subjected to the same boundary conditions.

Coming back to your original questions:

  • In the first general solution, the coeffs. $c_1$ and $c_2$ are integration constants, which are needed in order to fulfill the boundary conditions. They do not really have any physical meaning. However, if you impose the two boundary conditions stated above, the coeff. $c_1$ becomes $x_0$ and $c_2$ becomes $v_0/\omega$.
  • In the second general solution, the coeffs. $A$ and $\phi$ are the amplitude and phase of the oscillation. If we take the boundary conditions stated above, we obtain $A=x_0$ and $\phi = 0$. However, nothing prevents us from taking $v_0 \ne 0m/s$. In that case $A \ne x_0 = x(t=0s)$ and $\phi \ne 0$.
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c1 and c2 are [L]1, but what they represent?

Try writing the general solution like this:

$$x(t) = x(0) \cos(\omega t) + \frac{\dot x(0)}{\omega} \sin(\omega t)$$

Let's see if that works for $t = 0$:

$$x(0) = x(0) \cos(0) + \frac{\dot x(0)}{\omega} \sin(0) = x(0)$$


Now, take the time derivative and evaluate at $t = 0$:

$$\dot x(0) = -\omega x(0) \sin(0) + \omega \frac{\dot x(0)}{\omega} \cos(0) = \dot x(0)$$


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Your assumption with $c_1$ and $c_2$ is correct, and, stated by Semoi, $x(t)=c_1 \cos(t\omega)+c_2 \sin(t\omega)$ and $x(t)=Asin(t\omega + \phi)$ are equivalent and are acceptable general solutions to this case. $A$ is also $x_o$ in this case, yes, in that it is conventionally maximum displacement from equilibrium.

$\phi$ is simply the phase constant, which has to do with the initial conditions of the event happening. This is why you need initial conditions to find it out, but essentially, it's accounting for the fact that your pendulum, let's say, wasn't at a defined $x_o$ at time $t = 0$, so your sine curve wouldn't have to start at the $0$ point every time.

However, a better model to describe SHM if we want to consider $x(t)$ would be $x(t)=Acos(\omega t + \phi)$ rather than sine, since then the function starts at $A$, which would make sense since typically $x_0 = x = A$ if we make $t=0$ when you let the pendulum go after moving it from equilibirum. You'll notice that as Semoi said with $v_0 = 0\ m/s$ the equation reduces to that as well. However, this is just convention and you're perfectly fine using those formulae, but I recommend using cosine if you're measuring SHM in the traditional sense.

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