How Do Phase Differences Affect the Amplitude of Combined Waves?

[AndreAo]

Homework Statement

Determine the amplitude of the resultant wave when two sinusoidal waves having the same frequency and traveling in the same direction are combined, if their amplitudes are 3.0 cm and 4.0 cm and they differ in phase by $$\pi/2$$ rad

Homework Equations

y(x,t)=a*sin(kx-wt)
y(x,t)=b*sin(kx-wt+$$\phi$$)
$$\phi$$ phase difference
sin a + sin b = 2sin 1/2(a+b) . cos 1/2(a-b)

The Attempt at a Solution

Because the amplitudes are not the same couldn't solve it using the sum of sin.

 

[Redbelly98]

Welcome to Physics Forums

I think it would be useful to use the trig identity,

sin(a+b) = _____?​

where

a = kx-wt
b = φ​

 

[AndreAo]

Thanks

Using the principle of superposition:
y(x,t)=a*sin(kx-wt)+b*sin(kx-wt+$$\varphi$$)
Using sin(a+b)=sin a*cos b+sin b.cos a on the second sin of the expression above leads to:
y(x,t)=a*sin(kx-wt)+b*cos(kx-wt)
But I don't see a way to group sin and cos.

 

[Redbelly98]

Thanks

Using the principle of superposition:
y(x,t)=a*sin(kx-wt)+b*sin(kx-wt+$$\varphi$$)
Using sin(a+b)=sin a*cos b+sin b.cos a on the second sin of the expression above leads to:
y(x,t)=a*sin(kx-wt)+b*cos(kx-wt)

Uh, not quite. Let's look at just the sin(kx-wt + φ) term.

sin(A+B) = sinA*cosB + sinB*cosA is correct.

So what does sin(kx-wt + φ) equal? Again, use

A = kx-wt
B = φ

(I've changed a&b into A&B, since a&b were already defined as something else in the problem statement. I missed that before, sorry.)

 

[AndreAo]

Using just in sin(kx-wt+$$\phi$$):
sin(kx-wt+$$\phi$$) = sin (kx-wt)*cos $$\phi$$+ sin($$\phi$$)*cos (kx-wt)
cos $$\phi$$ = 0
sin $$\phi$$ = 1
so sin(kx-wt+$$\phi$$) = cos (kx-wt). What is wrong?

 

[Redbelly98]

Oh! I missed that φ=π/2, sorry about that.

So you were right before,

y(x,t)=a*sin(kx-wt)+b*cos(kx-wt)

Will have to think about this some more.

 

[ehild]

You want to replace the two terms by a single sine or cosine function of kx-wt, with amplitude A and phase beta.


y(x,t)=a*sin(kx-wt)+b*cos(kx-wt) = A sin (kx-wt + beta).

Apply the rule for sin (y+z) again. You will have sine and cosine of (kx-wt) on both sides of the equation. As this is an identity, it has to be true for all values of kx-wt, so the factor of the sine term on the right is "a" and that of the cosine term is "b". Then you have two equations for A and beta, solve for A.

ehild

 

[AndreAo]

The result before was:
y(x,t)=a*sin(kx-wt)+b*cos(kx-wt)
a*sin(kx-wt)+b*cos(kx-wt) = A sin(kx-wt+$$\beta$$)
Applying the rule on the right side of equation:
A[sin(kx-wt)*cos $$\beta$$+sin $$\beta$$*cos(kx-wt)]
So, Acos $$\beta$$ = a and Asin $$\beta$$ = b
A = a/cos $$\beta$$ and A = b/sin $$\beta$$
a/cos $$\beta$$ = b/sin $$\beta$$
tan $$\beta$$ = b/a
tan $$\beta$$ = 0.04/0.03
tan $$\beta$$ = 1.33
$$\beta$$ = arctan 1.33
$$\beta$$ = 0.92
A = a/cos 0.92 = 0.05
The answer is y(x,t)= 0.05sin(kx-wt+0.92) which is the book answer
Redbelly and ehild, thanks for helping!

 

[Redbelly98]

You're welcome!