Find the superposition of 2 waves, the ans is -6.02 cm, yet confused

[polarbearkids]

Homework Statement

The question asks us to find the superposition of 2 waves where y1=3cos(4x-1.6t) and y2= 4sin(5x-2t) and find the superposition of the waves y1 + Y2 at x=1 and t =0.5

Homework Equations

The Attempt at a Solution

Okay, what confuses me. is my teacher explained how to work with waves with the same amp, but not different and what do these equations give you. What does y1 mean? Is it the vertical distance?
The answer is not just plug in x and t, then add y1 + y2. I am so confused! please help! I know the waves are going the same way because of the sign but that is where my knowledge stops.

t is in s and x is in cm

And I got the answer from the back of the book.

 

[rude man]

Superposition is just addition. So just do y1 + y2 and substitute x = 1 and t = 0.5 in the sum.

If your teacher wanted you to simplify or othererwise re-express the sum he/she should have said so.

y1 and y2 are the orthogonal excursions of the H or E field. The wave travels in the x direction.

 

[Tanya Sharma]

Homework Statement

The question asks us to find the superposition of 2 waves where y1=3cos(4x-1.6t) and y2= 4sin(5x-2t) and find the superposition of the waves y1 + Y2 at x=1 and t =0.5

Homework Equations

The Attempt at a Solution

Okay, what confuses me. is my teacher explained how to work with waves with the same amp, but not different and what do these equations give you. What does y1 mean? Is it the vertical distance?
The answer is not just plug in x and t, then add y1 + y2. I am so confused! please help! I know the waves are going the same way because of the sign but that is where my knowledge stops.

t is in s and x is in cm

And I got the answer from the back of the book.

Hi polarbearkids...

Welcome to PF!

Just substitute x=1 and t=0.5 and do y1+y2 .You will get the correct answer.

y1 and y2 represent displacement of the particles due to the individual waves .Performing y1+y2 gives you the resultant displacement due to the two waves.

 

[polarbearkids]

Hi polarbearkids...

Welcome to PF!

Just substitute x=1 and t=0.5 and do y1+y2 .You will get the correct answer.

y1 and y2 represent displacement of the particles due to the individual waves .Performing y1+y2 gives you the resultant displacement due to the two waves.

Okay, I feel a little stupid and I was just frustrated! My question was more along these lines. What if they did not give us the x and t, but they gave us the phase shift as .25, so the question would be along the lines of y1=3cos(4x-1.6t) and y2= 4sin(5x-2t -.250) and we were asked to find the new amplitude.

the text gives the equation for a resultant wave is y = 2A cos(∅/2)sin(kx-wt+∅/2), and I know this part 2A cos(∅/2) is the amp. but I would suspect that would only be true if the amplitudes, wavelength and frequencies are the same. So how would your equation change to account for these different numbers? Is it simply the sum of the amplitudes? I thought about this. And if it was the sum of the amplitudes, the if you have the same amplitudes just one is negative and one is positive( like -3 and 3), then that A term would always equal zero. Yet, I know this cannot be true, because if there is a phase shift, then it amplitudes at that point would be different. So, could you please explain this to me. Hopefully you understand my confusion! I am bad at physics thinking!

 

[Tanya Sharma]

Okay, I feel a little stupid and I was just frustrated! My question was more along these lines. What if they did not give us the x and t, but they gave us the phase shift as .25, so the question would be along the lines of y1=3cos(4x-1.6t) and y2= 4sin(5x-2t -.250) and we were asked to find the new amplitude.

the text gives the equation for a resultant wave is y = 2A cos(∅/2)sin(kx-wt+∅/2), and I know this part 2A cos(∅/2) is the amp. but I would suspect that would only be true if the amplitudes, wavelength and frequencies are the same. So how would your equation change to account for these different numbers? Is it simply the sum of the amplitudes? I thought about this. And if it was the sum of the amplitudes, the if you have the same amplitudes just one is negative and one is positive( like -3 and 3), then that A term would always equal zero. Yet, I know this cannot be true, because if there is a phase shift, then it amplitudes at that point would be different. So, could you please explain this to me. Hopefully you understand my confusion! I am bad at physics thinking!

Amplitude is the maximum displacement of the particles of a wave.It is a positive quantity and is a constant for a wave.Different particles of a wave have same amplitude,but different displacements .In general ,the resultant amplitude of the superposition of waves is not the algebraic sum of the amplitudes of the two waves.

The displacement of the particle and the amplitude are two different things .Displacement is represented by 'y' and amplitude by 'A' .The displacements of the particles are different whereas the amplitude is same.

So,basically you are looking for superposition of waves where the amplitudes are different. Is it so?

 

[polarbearkids]

Amplitude is the maximum displacement of the particles of a wave.It is a positive quantity and is a constant for a wave. (- Thanks I forgot about that.

So,basically you are looking for superposition of waves where the amplitudes are different. Is it so?

But yes, that is what I am wondering about. What if the amplitudes are different. Even though, I strayed from this in my last post. Take the example of the lines of y1=3cos(4x-1.6t) and y2= 4sin(5x-2t -.250), how would you go about finding the new amplitude of the combined waves at any particular point. Would you use y = 2A cos(∅/2)sin(kx-wt+∅/2)? would this equation still be valid? if so, how would you find that A in that equation? and if not, how would you go about finding this new amplitude? with the old amplitudes being 3 and 4 respectively

 

[Tanya Sharma]

Okay...

Let the two waves be y₁=A₁cos(kx-ωt) and y₂=A₂cos(kx-ωt+∅)

y₁+y₂ = A₁cos(kx-ωt) + A₂cos(kx-ωt+∅)

=A₁cos(kx-ωt)+A₂[cos(kx-ωt)cos∅-sin(kx-ωt)sin∅]

=A₁cos(kx-ωt)+A₂[cos(kx-ωt)cos∅]-A₂sin(kx-ωt)sin∅

=[A₁+A₂cos∅]cos(kx-ωt)-A₂sin∅sin(kx-ωt)

Now put [A₁+A₂cos∅] =Rcosθ
and A₂sin∅ = Rsinθ

So now we have , y₁+y₂=Rcosθcos(kx-ωt)-Rsinθsin(kx-ωt)

y₁+y₂= Rcos(kx-ωt+θ) where R is the resultant amplitude .

R = √[(A₁+A₂cos∅)² + (A₂sin∅)²]

and θ = tan^-1[(A₂sin∅)/(A₁+A₂cos∅)]

Hope this helps

 

[polarbearkids]

Okay...

Let the two waves be y₁=A₁cos(kx-ωt) and y₂=A₂cos(kx-ωt+∅)

y₁+y₂ = A₁cos(kx-ωt) + A₂cos(kx-ωt+∅)

=A₁cos(kx-ωt)+A₂[cos(kx-ωt)cos∅-sin(kx-ωt)sin∅]

=A₁cos(kx-ωt)+A₂[cos(kx-ωt)cos∅]-A₂sin(kx-ωt)sin∅

=[A₁+A₂cos∅]cos(kx-ωt)-A₂sin∅sin(kx-ωt)

Now put [A₁+A₂cos∅] =Rcosθ
and A₂sin∅ = Rsinθ

So now we have , y₁+y₂=Rcosθcos(kx-ωt)-Rsinθsin(kx-ωt)

y₁+y₂= Rcos(kx-ωt+θ) where R is the resultant amplitude .

R = √[(A₁+A₂cos∅)² + (A₂sin∅)²]

and θ = tan^-1[(A₂sin∅)/(A₁+A₂cos∅)]

Hope this helps

Yes! This is exactly what I wanted. Thank you so much, this is starting to make sense.

Now, I have two final questions.

Will R = √[(A1+A2cos∅)^2 + (A2sin∅)^2] always be true or is it just when A2>A1? would R = √[(A1+A2cos∅)^2 + (A1sin∅)^2] if A2<A1. I don't think it would be, I just want to make sure.

and then finally, What angle does this give you θ = tan-1[(A2sin∅)/(A1+A2cos∅)]? like what information does that angle tell you, its the angle of __ in relation to the new wave. ??

Thank you!

 

[Tanya Sharma]

Yes! This is exactly what I wanted. Thank you so much, this is starting to make sense.

Now, I have two final questions.

Will R = √[(A1+A2cos∅)^2 + (A2sin∅)^2] always be true or is it just when A2>A1? would R = √[(A1+A2cos∅)^2 + (A1sin∅)^2] if A2<A1. I don't think it would be, I just want to make sure.

This is a general relation.

and then finally, What angle does this give you θ = tan-1[(A2sin∅)/(A1+A2cos∅)]? like what information does that angle tell you, its the angle of __ in relation to the new wave. ??

Thank you!

θ is the initial phase shift of the resultant wave .