When to use $x = A \sin (wt + p)$ and $x = A \cos (wt+p)$ during Simple Harmonic motion when we start our clock not between mean and extreme position?

Question

When to use $x = A \sin (wt + p)$ and $x = A \cos (wt+p)$ during Simple Harmonic motion when we start our clock when the object is between mean position and extreme position ?

I have been trying to study, and I know that when we start from extreme position, we use cosine, and sine for mean, but what if we start in between?

I know that both have a phase difference of 90 ' , but , if an initial phase is given , how to determine that it is p for x = A sin (wt + p) or = A cos (wt+p).

You can explain with the help of this problem. Thank you.

Answer 1

The simple harmonic oscillator is solved by the differential equation

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

This differential equation is second order, so it needs two initial conditions. This means that you likely need two constants. It turns out that there are multiple ways to write this solution (taking $\omega = \sqrt{k})$:

$$ x(t) = A \sin \omega t + B \cos \omega t $$

where $A, B$ are found by $x(0)$ and $\dot{x}(0)$,

$$ x(t) = C \sin ({\omega t + \delta_1}) $$

where $C, \delta_1$ are found by the initial state, or

$$ x(t) = D \cos (\omega t + \delta_2) $$

where $D, \delta_2$ are found by the initial state.

You may also see a solution written as $A e^{i \omega t}$, which can be related to the other solutions by taking the real part of Euler's formula.

These are all possible since the components share the same frequency and are only out of phase with each other. $\sin x$ and $\cos x$ are 90 degrees out of phase, but the additional phase factors $\delta$ make it so that either solution will work; you just have to solve for the constants with initial conditions.

Answer 2

This is an initial value problem. The general solutions to harmonic motion is either $$x(t)=A\sin(\omega t+\phi)$$ or $$x(t)=A\cos(\omega t+\phi)$$ or even $$x(t)=A\cos\omega t+B\sin\omega t$$ They are all equivalent they just give different values of $\phi, A$ or $B$. Pick the one you like most. To determine these constants you need initial values (you can also call them boundary conditions). You have two unknowns so you need two initial conditions. An example would be knowing that at $t=0$ the oscillator has position $x(0)=x_0$ and velocity $v(0)=v_0$. When you consider an extreme position you still have two initial values you just have $v(0)=0$. If we pick the $A\sin(\omega t+\phi)$ solution this gives the conditions $$\cases{A\sin\phi=x_0\\A\omega\cos\phi=v_0}$$ which in principle allows you to solve for $\phi$ and $A$. Does this answer your question?

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I realise I might have misread your question a bit. The equations that accompany the figure should be \begin{align} x(t)&=A\cos(\omega t+\phi)\\ y(t)&=A\sin(\omega t+\phi) \end{align} Why? At $t=0$ this describes a point at angle $\phi$: \begin{align} x(0)&=A\cos(\phi)\\ y(0)&=A\sin(\phi) \end{align} This follows from some trigonomotetry

If I understand correctly in the question they define $\phi$ as the angle the points makes at $t=0$ but I'm not sure since that would require more context.

Answer 3

Notice that $\cos(\theta) = \sin(\theta+90^o)$ so $\cos$ and $\sin$ functions have identical shapes with a phase difference of $90^o$ (or $\frac \pi 2$ radians). To describe SHM you can use either $x =A \cos (\omega t + p)$ or $x =A \sin (\omega t + p')$ where $p' = p + 90^o$ - the two descriptions are identical. For example, if you know that $x(0) = \frac 1 2 A$ and $x$ is decreasing at time $0$ then you can say either $x=A\cos( \omega t + 60^o)$ or $x=A\sin( \omega t + 150^o)$.

Answer 4

you have three equivalence solutions for SHM.

$$x_1(t)=A\sin(\omega\,t+p_1)\tag 1$$ $$x_2(t)=A\cos(\omega\,t+p_2)\tag 2$$ $$x_3(t)=B\sin(\omega\,t)+C\,\cos(\omega\,t)\tag 3$$

$x_2(t)=x_1(t)~$ for $~p_2=p_1+\frac{\pi}{2}$

expanding Eq. (1)

$$x_1=A \left( \sin \left( \omega\,t \right) \cos \left( p_{{1}} \right) + \cos \left( \omega\,t \right) \sin \left( p_{{1}} \right) \right) $$ and comparing the coefficients with Eq. (3) you get:

$$B=A\,\cos(p_1)~,C=A\,\sin(p_1)$$

so $~x_3(t)=x_1(t)$