Where must one use $x=A \cos \omega t$ and $x=A \sin \omega t$? [duplicate]

Question

Consider the following example:

A particle oscillates with simple harmonic motion in a straight line. In the first $t$ seconds, it travels a distance $a$. In the next $t$ seconds, it travels an additional $2a$ distance (in the same direction). Find the time period of oscillation.

For this question, the correct answer only comes when assuming $x=A\cos \omega t$. When using $x=A\sin \omega t$, the answer does not exist. Why?

Answer 1

From the specific wording of the question (I.e. it travels an additional 2'a' distance), I think you can start by imagining what simple harmonic motion looks like. The only time it will oscillate equal distances in equal time increments in the same direction is if it is oscillating around the center (whatever your ‘zero’ is) having started from one side and moving on to the other (as in it didn’t start from the center like sine does). The oscillatory solution of the simple harmonic oscillator that starts at a peak and moves to the other side is the cosine function. The way the question is worded however is sort of ambiguous as to if you can actually solve for the period because it doesn’t explicitly say this distance ‘a’ is the full amplitude.

If you are wanting to know why cosine and sine are the only solutions, it comes from solving the differential equation.

Edit: I reread the question and I think that it says “the first t seconds” means you are supposed to assume this is the beginning of oscillation. I had initially interpreted it as just meaning the beginning of measuring/observing the oscillation. In this case ‘a’ is the amplitude and it means the oscillation started from a peak, therefore only cosine will work as sine starts from 0.

Answer 2

Yes, Here you will only get the answer with using $x=A\cos{wt}$, Because your question mentioned that, In the first $t$ seconds the body moves distance $a$ and in the next $t$ seconds it moves distance $ 2a$. This is possible only if the body is left from an extreme position and moving towards the center. Because at the center, the body in SHM has the highest kinetic energy and hence will travel a greater distance in the same time interval. when left from the center the body in SHM will travel less distance in equal interval of time

In SHM, if the body starts moving from center, one uses $x=A\sin{wt}$ as at\ $ t=0, x=0.$ when the body starts moving from extreme position we use $x= A\cos{wt} $ as at $ t =\frac{\pi}{2}, x=0 $

In very general one uses $x=A\sin{wt}+B\cos{wt}$ and the initial condition tells you which one you have to pick up ($sin(x),\ cos(x) \ or\ combination\ of\ sin(x)\ and\ cos(x)$). for instance if you are leaving the particle from extreme then at $t=0\ x= C (say) $ (I know it will be B)$\ and\ at \ t=pi/2 , x=0 $ and that implies A=0.{see by putting $t=\frac{\pi}{2} $}

Answer 3

It is actually related to the initial conditions at $ t=0 $.

For example, let's see these two cases of a box connected to a spring.

First case:

|  _    _  |######|
|-/ \  / \_|######|
|    \/    |######|

^     ^       ^
|     |       |
Wall  Spring  Block

|-----------------|
x=0               x=A

We see that at $ t=0 $ the position is at $x=A$. Now we can use either cosine or sine as you like it wont matter , as long as you put the phase constant. I will use cosine as it is more common. $$x_{t=0} = A\cos(\omega t + \phi) = A\cos(\phi) = A \quad\therefore \phi =0, $$ where $\phi$ is the phase constant. We see that in this case $ x = A\cos(\omega t) $. Lets look at the second example.

Second Case:

||######|
||######|
||######|
 
 ^
 |
The spring is in between.

|-----------------|
x=0               x=A

We see that at $ t=0 $ the position is at $x=0$. Let's do as before. $$x_{t=0} = A\cos(\omega t + \phi) = A\cos(\phi) = 0 \quad\therefore \phi =90^{\circ}$$ we see that $x = A\cos(\omega t + 90^{\circ}) = A\sin(\omega t)$.