Answer:
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
