Where does the extra $\omega$ come in velocity of Simple Harmonic Motion?

Question

Position $x$ in a SHM is given by $x=A\space sin(\omega t+\phi)$.

Where $A$,$\omega$ and $\phi$ are Amplitude,Angular frequency and phase constant and are three constants respectively.

So,velocity of SHM is given by-

$$\frac {d}{dt}A\sin(\omega t+\phi)$$.

In my book it is given solving this it gives -$A\omega\space \cos(\omega t+\phi)$

Now,my question is where does the extra $\omega$ beside $A$ come from?

It would be greatly helpful if some one explains this to me.

Thanks for any help!!

Answer 1

By the chain rule of differentiation,

$$\frac {d}{dt}A\sin(\omega t+\phi)=\frac {d}{d(\omega t)}A\sin((\omega t)+\phi).\frac{d(\omega t)}{dt}=Acos(\omega t+\phi).\omega=A\omega cos(\omega t+\phi)$$

Answer 2

GoodDeeds has already provided a sufficient answer, but I want to explain why you should have expected the velocity to be proportional to $A\omega$. This answer might be more accessible to someone who hasn't had calculus.

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$A$ has the dimension of length, i.e., it can be measured in meters. The velocity is a measure of the change in displacement over time. The formula for the velocity can't just depend on $A$, because $A$ can't provide the necessary dimension of time.

The only other parameter we have which describes the motion is the frequency $\omega$. This frequency has dimensions of inverse time, i.e., $\omega$ can be measured in units of $s^{-1}$.

Because velocity requires both time and length it must depend on a combination of $A$ and $\omega$. In particular the product $A\omega$ has the correct dimensions of length divided by time, i.e., $A\omega$ can be measured in $m/s$. From this we conclude that the velocity should have the form,

$$ v = A\omega \left( \text{Something} \right).$$

Answer 3

It's because of the derivative of a "function of a function". Indeed it's like you had (forgetting about $A$ for the moment)

$$\sin[f(t)]$$

where $f(t) = \omega t + \phi$

So when you derive such a function you get:

$$\frac{\text{d}}{\text{d}t}\sin[f(t)] = \cos[f(t)]\cdot f'(t)$$

your $f'(t) = \omega$ thence the result:

$$\frac{\text{d}}{\text{d}t}A\sin[f(t)] = A\omega\cos(\omega t + \phi)$$