Question

Two traveling sinusoidal waves are described by the wave functions

y1 = 4.75 sin [π(4.30x −  1115t)]

y2 = 4.75 sin [π(4.30x −  1115t − 0.250)]

where x, y1, and y2 are in meters and t is in seconds.

(a) What is the amplitude of the resultant wave function 

y1 + y2?

(b) What is the frequency of the resultant wave function?

Answer 1

Use the trig identity:

sinA+sinB=2sin(½(A+B))cos(½(A-B))

When B=A+C, this becomes 2sin(A+½C)cos(½C).

A=π(4.30x-1115t) and C=-0.250π=-π/4.

We can incorporate the common amplitude 4.75 into this to give us:

9.5sin(A+½C)cos(π/8) because cos(-π/8)=cos(π/8).

(a) The new amplitude is 9.5cos(π/8)=8.7769 approx.

(b) Frequency=F=1/Period=1/T, where T can be found by examining the sine argument.

So, π(4.30x-1115t-0.125)=2πn where n is an integer makes sine=0.

That is, 4.30x-1115t-0.125=2n. Initially at (x₀,t₀), n=0 and t₀=(4.30x₀-0.125)/1115.

After the first cycle, t₁=(4.30x₁-2.125)/1115. T=t₁-t₀ and F=1/(t₁-t₀).

F=1115/(4.30(x₁-x₀)-2). If x₀=0, then we can write x₀ simply as x,

F=1115/(4.30x-2).

Answer 2

@ people in the future looking at this: that other answer is so overcomplicated it makes me physically ill. just remember to distribute in that pi and make sure youre in radians.

a) amplitude = 2Acos(phi/2) m

2(4.75)cos(0.25pi*0.5) = 8.777 m

b) frequency = w/(2pi) Hz

1115pi/2pi = 557.5 Hz

those are standard equations for superposition of sinusoidal waves, their derivation is gonna be in your book.