# B Simple Harmonic Motion: why sin(wt) instead of sin(t)?

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1. Sep 15, 2018

### thebosonbreaker

Hello,
I have recently been introduced to the topic of simple harmonic motion for the first time (I'm currently an A-level physics student). I feel that I have understood the fundamental ideas behind SHM very well. However, I have one question which has been bugging me and I can't seem to find a valid answer.
I will consider the example of a simple pendulum which has been set in motion and therefore oscillates about a fixed equilibrium position.
If the displacement of the pendulum bob is considered as a function of time, then the graph of x/t is analogous to a sine curve (assuming that the bob is released from a point of maximum displacement - since I believe that starting from the equilibrium position would produce a cosine curve [please correct me if I'm wrong here])
The graph will have the equation x = Asin(wt). Now I try to break this down in order to understand why this equation is true for SHM.
Firstly, as I said the variation of x with t produces a sine curve, explaining why X is a function of sin(t). I'm fine with that. Next, I understand that A (the amplitude or max. displacement) is a coefficient on the outside because it has the effect of 'stretching' the sine curve (since the bob oscillates between positions of max. displacement either side.) Again, that all makes sense. However, what I do not understand is why sin(wt) [I know omega is the letter used but I only have w on my keyboard] is used instead of sin(t). It is time which is plotted on the horizontal axis, so surely the y axis represents displacement (x) and the x axis represents time (t).
Why is the equation not:
x = Asin(t)?

If somebody could clear this up for me it would be greatly appreciated.

2. Sep 15, 2018

### PeroK

Not all SHM has the same period. If we are measuring time in seconds, then $\sin(t)$ would imply a period of $2\pi$ seconds. Whereas, for $\sin(\omega t)$ the period is $\frac{2\pi}{\omega}$, which covers the general case where $\omega$ determines the period (or vice versa).

For example, if the period of the pendulum is 5 seconds, then $\omega = \frac{2\pi}{5}$.

3. Sep 15, 2018

### Staff: Mentor

If you have $y=A\cdot \sin(x)$ then you have determined a set of certain waves by varying the amplitude. But you can also vary the period, and whether the graph must contain $(0,0)$. So varying the period gives you $y=A\cdot \sin(\omega \cdot t)$ and a translation out $(0,0)$ an additional constant shift $C$, so all in all $y=A\cdot \sin (\omega \cdot t) + C$ defines the most general class of waves, if neither amplitude nor period varies in time. We could certainly adjust the coordinate system to make $A=\omega = 1$ and $C=0$, but what to do if a second wave is considered in parallel?

4. Sep 15, 2018

### robphy

The argument of the sine function must be radians (or dimensionless).
So, if t has units of time (e.g. sec), then sin(t) doesn't make sense.
Indeed, $\omega$ has units of rad/sec... so that $\omega t$ has units of radians (or is dimensionless).

Side comment:
Similarly, the argument of log() and exp() must be dimensionless.
In the stat mech class I am teaching, I complained about the textbook integrating $\int^{V_{b}}_{V_{a}}\frac{dV}{V}=\ln V_{b} - \ln V_{a} ,$ where $V$ is a volume.
It should be $\int^{V_{b}}_{V_{a}}\frac{dV}{V}=\ln \frac{V_{b}}{V_{a}} .$
If you really want to write a difference then one should write
$\int^{V_{b}}_{V_{a}}\frac{dV}{V}=\ln \frac{V_{b}}{V_{ref}} - \ln \frac{ V_{a}}{V_{ref}} .$, where $V_{ref}$ is any nonzero reference volume.