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

Amplitude of the resultant wave = 15.72 cm

Step-by-step 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