These two waves travel along the same string: y1 = (4.17 mm) sin(2.24?x - 300?t), y2 = (5.96 mm) sin(2.24?x - 300?t + 0.727?rad). What are (a) the amplitude and (b) the phase angle (relative to wave 1) of the resultant wave? (c) If a third wave of amplitude 5.20 mm is also to be sent along the string in the same direction as the first two waves, what should be its phase angle in order to maximize the amplitude of the new resultant wave?

a) Amplitude is 9.49 mm

b) ∅ = 0.43056 rad = 24.7°

c) phase angle = 2.24x - 300t + 0.43056 rad

Explanation:

Given that;

A1 = y1 = (4.17 mm) sin(2.24x - 300t)

A2 = y2 = (5.96 mm) sin(2.24x - 300t + 0.727rad)

now phase difference between y1 and y2 is;

Δ∅ =  (2.24x - 300t + 0.727rad) - (2.24x - 300t) = 0.727 rad =  41.65°

a)  the amplitude

Amplitude A = √( A1² + A2² + 2A1A2cosΔ∅)

we substitute

A = √( (4.17)² + (5.96)² + (2 × 4.17 × 5.96 × cos(41.65) )

A = √( 17.3889 + 35.5216 + ( 49.7064 × 0.7472 )

A = √(52.9105 + 37.1406 )

A =  √90.0511

A = 9.49 mm

Therefore, Amplitude is 9.49 mm

b) the phase angle (relative to wave 1) of the resultant wave;

tan∅ = A2sinΔ∅ / ( A1 + A2cosΔ∅)

we substitute

tan∅ = 5.96sin41.65 / ( 4.17 +  5.96cos41.65)

tan∅ = 3.96088 / ( 4.17 + 4.4534)

tan∅ = 3.96088 / 8.6234

tan∅ = 0.4593

∅ = tan⁻¹ ( 0.4593 )

∅ = 0.43056 rad = 24.7°

c)

For the third wave, maximum amplitude. it should be in the direction of resultant of A1 and A2

so, phase angle in order to maximize the amplitude of the new resultant wave will be;

phase angle = 2.24x - 300t + 0.43056 rad

answer: a. same unit

explanation: