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For simple harmonic motion (SHM), I am aware you can start of using either sine or cosine, but I am a bit confused as to when you would start off with sine rather than cosine. I know that a sine graph starts at $y=0$ and a cosine graph starts at $y=1$. So therefore, I would say you use sine for equilibrium positions?

However, I came across a question asking to write down the equations for position, velocity and acceleration of a particle starting from rest at time $t=0$, then undergoing SHM with maximum amplitude 0.2 m and period 5 sec.

I worked out the angular frequency to be $2\pi/5$ from the period formula. And then used the position formula to be of the form with sine and differentiated to get cosine velocity equation, etc. However, the answer says I should have started with cosine and I am now unsure when I should start with sine or cosine.

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  • $\begingroup$ Either answer is correct. If you would normally use sin and you used cos, there should be a phase correction of 90 deg in your cos argument. $\endgroup$ – David White Nov 30 '18 at 16:06
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The function $x(t) = A \sin \omega t$ starts from zero with maximum speed, while the function $x(t) = A \cos \omega t$ starts from $x=A$ (the amplitude) with zero speed, and starts to move towards $x=0$. Starting from rest doesn't imply starting at the equilibrium position: if you start from rest at $x=0$, nothing moves, so this is not an interesting solution! For a harmonic oscillator, starting from rest means starting at the maximum value of $x$, so $\cos \omega t$ is the appropriate solution.

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"a particle starting from rest at time t=o" This is the key to why you start with cosine. So you know that instantaneous velocity is the derivative at that point. Look at a sine graph at t=0 there is a positive gradient thus the particle doesn't start at rest with sine. But with cosine at t = 0 the gradient is zero thus the particle is initially at rest.

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Since the return force in general can be taken to be $F(x)=-kx$, Newton's second law renders the following differential equation: $$-kx=m\frac{d^2x}{dt^2}$$ whence $$\boxed{\frac{d^2x}{dt^2}+\omega^2x=0}$$ where I substituted $\omega\equiv \sqrt{k/m}$.

A solution to this equation in its most general form is $$x=A\sin\omega t +B\cos\omega t$$ where $A,B$ are constants. A so-called initial condition such as $x(t=0)=0$ will rule out the cosine solution, since $x(0)=B=0$ according to the condition. Another one such as $x'(0)=a$ (meaning the velocity be $v=x'(t)=a$ at $t=0$) would then give you $A=a/\omega$ which completely determines $x(t)$ as $$\boxed{x(t)=(a/\omega)\sin\omega t}$$

Of course, these initial conditions can't be chosen arbitrarily for a specific problem. They're always "imposed" by the problem itself.

Two completely equivalent general solutions, however, are $x_1(t)=A\cos(\omega t+\phi_1)$ and $x_2(t)=B\cos(\omega t+\phi_2)$ where $\phi$ is called a phase constant. These can be taken to be separate general solutions to the problem - they don't need to be added! Initial conditions will determine the two constants $A$ and $\phi_1$ or $B$ and $\phi_2$.

For example, take $x_1(t)=A\cos(\omega t+\phi_1)$. $x_1(0)=0$ renders $A\cos\phi_1=0$. Setting $A=0$ would be boring, because then $x_1(t)$ would simply be zero. Thus, we need $\phi_1$ to be any odd multiple of $\pi/2$, since that's where the cosine vanishes. $\phi=\pi/2$ will do the job. $x_1'(0)=a$ would give you $$-A\omega\sin(\pi/2)=a,$$ whence $A=-a/\omega$. The solution to our problem then becomes $$\boxed{x(t)=(-a/\omega)\cos(\omega t + \pi/2)}$$ where I dropped the subscript one. This is completely equivalent to $x(t)=(a/\omega)\sin\omega t$, since $$\cos(\omega t+\pi/2)=\cos(\pi/2 -(-\omega t))=\sin(-\omega t)=-\sin\omega t.$$ Plugging this into our second solution $$x(t)=(-a/\omega)\cos(\omega t + \pi/2),$$ we again get $$x(t)=(a/\omega)\sin\omega t.$$

This should demonstrate that when done carefully, any two different mathematical approaches to the same problem must render the same solution.

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