Finding the period of a sinusoid

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tade
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Let's have y = sin(k(a+vt))*sin(ωt)

where k, ω, a and v are all positive real numbers.

What is the period of this sinusoid in terms of k, ω, a and v?
 
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That's a good question - what have you tried?
i.e. what happens if you try to turn your product of sinusoids into a single trig function?

This is: ##A\sin k(x+vt)## where ##A=\sin\omega t## considered at point ##x=a## right?
Presumably ##\omega \neq kv## in this case?

In which case, you have an equation of form: $$y(t)=\sin(\omega_1 t + \phi)\sin(\omega_0 t) $$
 
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But it still looks the same though.


I believe the value of a doesn't matter.

Let's try sin(kvt)*sin(ωt) which simplifies to

sin(δt)*sin(ωt)


Now we only have two variables.
 
If sin(δt+δj)*sin(ωt+ωj) = sin(δt)*sin(ωt)

Then j should be equal to nT, where n is an integer and T is a constant based on δ and ω. The period of the wave.
 
The value of a just affects the relative phase between the two sinusoids.
What I suggested with the breakdown was that you treat the sin(wt) as the amplitude of the sin(k(x+vt)) traveling wave. What is happening?

Since you are only looking at the oscillations at one point in space, you are just multiplying sine waves together like you've shown: sin(δt)*sin(ωt) $$y(t)=\sin\delta t \sin\omega t$$ ... basically.

What sort of shape is that wave?

Do you know about beats?
Do you know about amplitude modulation?
 
If δ and ω are both integers, then j = n*2∏
 
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I don't know much about AM, but I do know that
$$y(t)=\sin\delta t \sin\omega t$$
is a form of AM
 
You should google the key words - "beats" as well.

The resulting waveform is not a simple sinusoid - so you have to figure out how the concept of a period applies here. What is correct depends on what you need the period for. How does this question come up?

i.e. If you need the time before the pattern starts to repeat, then the phase factor will be important too.
You should experiment by plotting the waves for different values of the parameters and see how it works.