# Amplitude of a resultant wave

1. Nov 5, 2009

### AndreAo

1. The problem statement, all variables and given/known data
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

2. Relevant 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)
3. The attempt at a solution
Because the amplitudes are not the same couldn't solve it using the sum of sin.

2. Nov 6, 2009

### Redbelly98

Staff Emeritus
Welcome to Physics Forums

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

sin(a+b) = _____?​

where

a = kx-wt
b = φ​

3. Nov 6, 2009

### 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.

4. Nov 6, 2009

### Redbelly98

Staff Emeritus
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.)

5. Nov 6, 2009

### 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?

6. Nov 6, 2009

### Redbelly98

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

So you were right before,

7. Nov 7, 2009

### 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

8. Nov 7, 2009

### 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