Question

Two waves are traveling in the same direction along a stretched string. The waves are 45.0° out of phase. Each wave has an amplitude of 9.00 cm. Find the amplitude of the resultant wave.

Answer 1

Answer:

Amplitude of the resultant wave = 15.72 cm

Explanation:

If two identical waves are traveling in the same direction, with the same frequency, wavelength and amplitude; BUT differ in phase the waves add together.    

A = 9cm (amplitude)

φ = 45 (phase angle)

The two waves are y1 and y2

y =  y1 + y2

where y1 = 9 sin (kx - ωt)    

and   y2 = 9 sin (kx - ωt + 45)

y       = 9 sin(kx - ωt) + 9 sin(kx - ωt + 45)  =  9  sin (a)   +   9  sin (b)

where a =  (kx - ωt)

abd b = (kx - ωt + 45)

Apply trig identity: sin a + sin b = 2 cos((a-b)/2) sin((a+b)/2)

A  sin (   a    )   +   A  sin (     b    ) = 2A cos((a-b)/2) sin((a+b)/2)

We have that

9  sin (   a    )   +   9  sin (     b    ) = 2(9) cos((a-b)/2) sin((a+b)/2)

= 2(9) cos[(kx - wt -(kx - wt + 45))/2] sin[(kx - wt +(kx -wt +45)/2]

y       = 2(9) cos (φ /2) sin (kx - ωt + 45/2)

The resultant sinusoidal wave has the same frequency and wavelength as the original waves, but the amplitude has changed:  

Amplitude equals 2(9) cos (45/2) = 18 cos (22.5°) = 18 * -0.87330464009

= -15.7194835217 cm ≅ 15.72 cm

since amplitudes cannot be negative our answer is 15.72 cm