Sine for SHM but Cosine for Spring-Mass System, where's the catch?

Question

Normally (correct me if I'm wrong) we have that Simple Harmonic Motion (SHM) is of the form $$y(t) = A\sin\left(\frac{2\pi t}{T}\right) = A\sin(\omega t)$$ where $A$ is the amplitude, $t$ is the independent variable of time and $T$ is the period of the motion and $\omega$ is the angular speed. (And anyway, angular speed or angular frequency?)

The spring-mass system on a frictionless plane should be SHM. However we normally find $$x(t) = A\cos\left(\sqrt{\frac{k}{m}}t\right)$$ when the spring is horizontal. Where $k$ is the spring coefficient, $m$ the mass and we have $\omega = \sqrt{\frac{k}{m}}$.

How come the equation of motion is given by cosine and not sine when Spring-Mass System on a frictionless plane is SHM and SHM should be described by a sine function?

Here's MIT Physics Lecture on SHM, which I find confusing as well about this little catch.

Now, my only understanding of it coming from the pdf is that the above equation with cosine holds when out initial conditions are that we displace the mass of an amount $A$ and we give no initial velocity to the mass.

Then is there a time in a spring-mass system (maybe with friction?) where we describe $x(t)$ (measured from the equilibrium position) as a sine function? If so, when? And if that is the case, how do we distinguish when to use the cosine function and the sine function?

@Blauelf thanks I just found it easier to write, but I'll edit it! — Euler_Salter, Jan 03 '17 at 14:35
The catch is in the initial conditions, which can induce different phases. — , Jan 03 '17 at 14:37
@Euler_Salter It could be swapped other way also, dude..you have the liberty to choose any boundary condition. Depends on whether you want to start at maximum value or zero. — Narasimham, Jan 03 '17 at 14:39
Right, Essentially they are same phenomenon. Initial or boundary condition are a matter of choice for the phase.. When you start your bike or car do you care to check if the air valve is at topmost or bottommost position? — Narasimham, Jan 03 '17 at 14:53

score 3 · Accepted Answer · answered Jan 03 '17 at 14:32

3

Sine and cosine are essentially the "same" function, one is just a 90 degrees phase shift to the other. The reason why cosine is chosen, is likely due to conventions in physics (satisfying initial conditions, as one criterion). Recall that the solution to the harmonic oscillator (no friction) $y''+ \beta y=0$ is given by $y(t)=Acos(\omega t)+Bsin(\omega t)$.

Recall that the ODE is linear, so by linearity (sometimes referred to as "superposition principle"), any linear combination will also work as a solution, in particular, when $A$ or $B$ are $0$, and this works out because we can always phase shift $sin(\omega t)$ to $cos(\omega t + \phi)$ and vice-versa.

answered Jan 03 '17 at 14:32

Kernel_Dirichlet

1,683

this makes so much sense! So basically the superposition principle here works because when we add the cosine solution to the sine solution, we are actually just adding a shifted sine to another sine? – Euler_Salter Jan 03 '17 at 14:40
Also, in your equation, do we have $\sqrt{\beta} = w $ ? – Euler_Salter Jan 03 '17 at 14:40
exactly! That's pretty much what's going on. Also, the linearity/superposition principle works for ANY linear ODE (and even PDE). The only catch is that the details and intuition become not as straightforward as with the classic mass on a spring. Also yes, $\sqrt{\beta}=\omega$ – Kernel_Dirichlet Jan 03 '17 at 14:43
perfect! Haven't to encountered PDEs just yet, but I'll keep it in mind, thank you! – Euler_Salter Jan 03 '17 at 14:44

Sine for SHM but Cosine for Spring-Mass System, where's the catch?

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