Voltage Frequency: Calculating wt

[suv79]

how can i work out the frequency of the voltage V1= 3sin(wt) and V2= 2cos(wt) ?

 

[f95toli]

I suspect this should be in the homework section...But...

Do you know how to write the expression for a sine wave with a frequency f?

 

[suv79]

W=2 pi f

 

[suv79]

i think that the wt is the same for both voltages, so the frequency will be the same,
i have added the two voltages together, which i believe can only be done if the frequency are the same.

 

[suv79]

but i am not sure how to verify it

 

[berkeman]

how can i work out the frequency of the voltage V1= 3sin(wt) and V2= 2cos(wt) ?

i think that the wt is the same for both voltages, so the frequency will be the same,
i have added the two voltages together, which i believe can only be done if the frequency are the same.

Welcome to the PF.

Yes, the 2 frequencies are the same. But you can add signals even if their frequencies are different.

Can you please type out the full question that you are trying to answer?

 

[suv79]

V1= 3sin(wt) and V2= 2cos(wt)
if
V3 = V1 + V2

find the expression of V3 in a sine waveform : V3 = Rsin(wt+α)

and verify that the resultant voltage V3 is in the same frequency as V1 and V2

 

[suv79]

∝=arctan(2/3)
∝=33.69°

R=√(3^2+2^2 )
R=3.61




V_3=V_1+V_2
3.61 sin(ωt+33.69°)=3 sin(ωt)+2 cos(ωt)

 

[suv79]

so i don't understand how i can verify its the same

 

[UltrafastPED]

First step: convert cosine to sine - they are out of phase.

Then see http://www.kwantlen.ca/science/physics/faculty/mcoombes/P2421_Notes/Phasors/Phasors.html

 

[suv79]

i don't know how do that

 

[UltrafastPED]

Look up trig identities: http://www.sosmath.com/trig/Trig5/trig5/trig5.html

 

[suv79]

cos(∏/2-u)=sinu

 

[suv79]

The frequency has not affected as the angular velocity is the same. ?

 

[UltrafastPED]

The frequency is the same for both terms, but they have different amplitudes, and different phases.

If you make a plot - plot each term, and their sum - three curves. Use different colors or symbols ... let w=2pi, and run the time from -2/2pi to +6/2pi to see a few cycles.

 

[suv79]

i just need to know that if the frequency are the same on V1+V2=V3

 

[gneill]

When you have an expression like:

$A sin(θ) + B cos(θ)$

which you want to reduce to a single sin or cos expression, there's a handy trick which makes it amenable to using your trig identities.

Since you can multiply any expression by 1 and leave it unchanged, multiply through by:
$$1 = \frac{\sqrt{A^2 + B^2}}{\sqrt{A^2 + B^2}}$$
Leave the numerator outside the expression, but take the denominator and place it under the A and B:
$$\sqrt{A^2 + B^2}\left(\frac{A}{\sqrt{A^2 + B^2}} sin(θ) + \frac{B}{\sqrt{A^2 + B^2}} cos(θ) \right)$$

Now $\frac{A}{\sqrt{A^2 + B^2}}$ and $\frac{B}{\sqrt{A^2 + B^2}}$ can be associated with the sin and cos of some other angle, say ∅. You can make either one sin(∅ )and the other cos(∅). Choose which is which by looking at your "angle sum" trig identities and make the overall pattern match either sin(θ ± ∅) or cos(θ ± ∅) as you wish.

 

[UltrafastPED]

i just need to know that if the frequency are the same on V1+V2=V3

Yes, the frequencies of the two terms are the same. You have been told this several times ...
For a refresher see: http://www.physicsclassroom.com/class/waves/

 

[Electest]

I am currently working my way through this question and can fully understand expressing the formulas similar to that of post 17 to reach V3= Rsin(wt+a). However the question suggests that both compound and double angle formulas be used and my method only seems to cover compound.

Also would I obtain a correct value of 'a' from V3= Rsin(wt+a) by using arctan2/3 = 0.588 rads or 33.69 degrees?

Thanks

 

[suv79]

I will have a look at my notes. I can't remember the question

 

[KEEPitSIMPLES]

why are you all finding the angle via ⅔? to find arctan is it not sin/cos? therefore this would be 3/2? giving the angle 56.31deg not 33.69. I have it remembered and in my notes that it is this way around, help as to why it has changed for this question would be greatly appreciated.

 

[gneill]

why are you all finding the angle via ⅔? to find arctan is it not sin/cos? therefore this would be 3/2? giving the angle 56.31deg not 33.69. I have it remembered and in my notes that it is this way around, help as to why it has changed for this question would be greatly appreciated.

It depends on whether you want the result to be expressed as a sin function or a cos function. The trig identity used to form the sum will be either:

$sin(θ + φ) = cos(φ) sin(θ) + sin(φ) cos(θ)$
or
$cos(θ - φ) = cos(φ) cos(θ) + sin(φ) sin(θ)$

See also post#17 in this thread to see how $cos(φ)$ and $sin(θ)$ are formed from the amplitude coefficients of the original voltage functions.

 

[KEEPitSIMPLES]

thats great, thank you