A plane progressive wave on a water surface is given by the equation y=2sin2pi(100t-x/30) where x-is the distance covered in a time t, x, y and t are cm and second respectively. Find (i) the wavelength and frequency of the wave motion. (ii) the phase difference between two points on the water surface that are 60cm apart.
wavelength is 30m
frequency is 100 Hz
phase difference is 540 degrees
please
To find the wavelength and frequency of the wave motion, we can start by comparing the given equation with the general equation of a plane progressive wave:
y = A sin (kx - ωt + φ)
Comparing the two equations, we can determine the values of A, k, ω, and φ from the given equation:
A = 2
k = 2π/λ (where λ is the wavelength)
ω = 2πf (where f is the frequency)
φ = 0 (since there is no initial phase mentioned in the given equation)
(i) Wavelength (λ):
From the equation, we have k = 2π/λ.
Comparing this with the equation above, we can see that k = 2π/30, so by solving this equation for λ, we get:
2π/λ = 2π/30
Dividing both sides by 2π:
1/λ = 1/30
Now, solving for λ:
λ = 30 cm
So, the wavelength of the wave motion is 30 cm.
(ii) Phase Difference (Δφ):
The phase difference (Δφ) between two points on the wave can be calculated using the formula:
Δφ = 2πΔx / λ
where Δx is the distance between the two points.
In this case, Δx = 60 cm. Substituting the values into the formula, we get:
Δφ = 2π(60) / 30
Simplifying the expression:
Δφ = 4π radians
So, the phase difference between the two points on the water surface that are 60 cm apart is 4π radians.