Two waves traveling on a string in the same direction both have a frequency of 92 Hz, a wavelength of 0.19 m, and an amplitude of 0.32 m. What is the amplitude of the resultant wave if the original waves differ in phase by pi /3 rad?
What is the phase difference between the two waves if the amplitude of the resultant wave is 0.25 m?
To find the amplitude of the resultant wave when the original waves have a phase difference of π/3 radians, we can use the concept of superposition. Superposition states that when two waves meet, their displacements add up to form a resulting wave.
Let's call the amplitude of the first wave A1 and the amplitude of the second wave A2. The resulting wave, with amplitude Ar, can be calculated using the formula:
Ar = sqrt(A1^2 + A2^2 + 2*A1*A2*cos(Δϕ))
Where Δϕ is the phase difference between the two waves.
Given that both waves have the same frequency, wavelength, and amplitude, we can say A1 = A2 = 0.32 m.
Substituting the given values into the formula, we have:
Ar = sqrt((0.32)^2 + (0.32)^2 + 2*(0.32)*(0.32)*cos(π/3))
Ar = sqrt(0.1024 + 0.1024 + 0.2048*cos(π/3))
Ar = sqrt(0.2048 + 0.2048 + 0.2048 * (1/2))
Ar = sqrt(0.6144 + 0.1024)
Ar = sqrt(0.7168)
Ar ≈ 0.85 m
Therefore, the amplitude of the resultant wave is approximately 0.85 m.
Now, let's find the phase difference between the two waves when the amplitude of the resultant wave is 0.25 m.
Using the same formula as above, we can rearrange it to solve for the phase difference Δϕ:
cos(Δϕ) = (Ar^2 - A1^2 - A2^2) / (2 * A1 * A2)
Substituting the given values:
cos(Δϕ) = (0.25^2 - (0.32)^2 - (0.32)^2) / (2 * 0.32 * 0.32)
cos(Δϕ) = (0.0625 - 0.1024 - 0.1024) / (2 * 0.32 * 0.32)
cos(Δϕ) = -0.1423 / 0.2048
cos(Δϕ) ≈ -0.6953
Taking the inverse cosine of both sides:
Δϕ ≈ arccos(-0.6953)
Using a calculator, we find:
Δϕ ≈ 2.495 radians
Therefore, the phase difference between the two waves is approximately 2.495 radians.