Two sinusoidal waves are moving through a medium in the positive x-direction, both having amplitudes of 7.00 cm, a wave number of k=3.00m-1, an angular frequency of ω=2.50s-1, and a period of 6.00 s, but one has a phase shift of an angle ϕ=π12rad. What is the height of the resultant wave at a time t=2.00s and a position x=0.53m?

0.99 m

Explanation:

Parameters given:

Amplitude, A = 7.00cm

Wave number, k = 3.00m^-1

Angular Frequency, ω = 2.50Hz

Period = 6.00 s

Phase, ϕ = π/12 rad

Note: All parameters are the same for both waves except the phase.

Wave 1 has a wave function:

y1 (x, t) = Asin (kx - ωt)

y1 (x, t) = 7sin (3x - 2.5t)

Wave 2 has a wave function:

y2 (x, t) = Asin (kx - ωt + ϕ)

y2 (x, t) = 7sin (3x - 2.5t + π/12)

π is in radians.

When Superposition occurs, the new wave is represented by:

y (x, t) = 7sin (3x - 2.5t) + 7sin (3x - 2.5t + π/12)

y (x, t) = 7[sin (3x - 2.5t) + sin (3x - 2.5t + π/12) ]

Using trigonometric function:

sin (a) + sin (b) = 2cos[ (a - b) / 2]sin[ (a + b) / 2]

Where a = 3x - 2.5t, b = 3x - 2.5t + π/12

We have that:

y (x, t) = (2*7) [cos (π/24) sin (3x - 2.5t + π/24) ]

Therefore, when x = 0.53cm and t = 2s,

y (x, t) = (2*7) [cos (π/24) sin{ (3*0.53) - (2.5*2) + π/24}]

y (x, t) = 14 * 0.9914 * 0.0713

y (x, t) = 0.99 m

The height of the resultant wave is 0.99cm