The waves emitted from a source A is given by
y1 = 6 sin π(20t – 0.5x)
where y1, x and t are expressed in units of meter and seconds.
(a) Determine the wavelength and the frequency of the radio waves. At the same time,
source B generates waves y2 that is identical to y1 except that there is a phase
difference of π/3
(b) Write down the wave equation for y2
If y2 were superimposed to y1
(c) what is the phase constant and the amplitude of the resultant wave?
(d) write down the equation of the resultant waves.
To determine the wavelength and frequency of the waves emitted from source A, we can compare the equation of the wave to the general wave equation:
y = A sin(2πft - kx + ϕ)
In this equation, A represents the amplitude, f represents the frequency, k represents the wave number, and ϕ represents the phase constant.
Comparing the given equation y1 = 6 sin π(20t – 0.5x) to the general wave equation, we can extract the values:
Amplitude (A) = 6
Frequency (f) = 20/2π
Wave number (k) = 0.5
Phase constant (ϕ) = 0
(a) To find the wavelength (λ), we can use the formula:
Wave speed = Frequency × Wavelength (v = fλ)
Since the wave speed is not mentioned, we can assume it to be the speed of light in a vacuum, which is approximately 3 × 10^8 m/s.
We can rearrange the equation to solve for the wavelength:
wavelength (λ) = wave speed (v) / frequency (f)
= (3 × 10^8 m/s) / (20/2π Hz)
= (3 × 10^8 m/s) / (10/π Hz)
= (3 × 10^8 m/s) × (π/10 Hz)
= 9.42 × 10^7 m
Therefore, the wavelength of the radio waves emitted from source A is approximately 9.42 × 10^7 meters.
(b) To determine the wave equation for y2, we can use the equation:
y2 = A sin(2πft - kx + ϕ)
Given that there is a phase difference of π/3, the phase constant (ϕ) for y2 would be π/3. Thus, the wave equation for y2 would be:
y2 = 6 sin π(20t – 0.5x + π/3)
(c) To find the phase constant and amplitude of the resultant wave when y2 is superimposed with y1, we need to consider the interference between the two waves. If the waves are superposed constructively, the resultant wave will have a maximum amplitude. If they are superposed destructively, the resultant wave will have a minimum amplitude.
Given that the waves y1 and y2 are identical except for a phase difference of π/3, they will superpose constructively. This means that the phase constant (ϕ) of the resultant wave will be the same as the phase constant of y1, which is 0, and the amplitude will be the sum of the amplitudes of y1 and y2, which is 6.
(d) The equation of the resultant wave when y2 is superimposed with y1 can be obtained by adding the two waves:
Resultant wave equation = y1 + y2
= 6 sin π(20t – 0.5x) + 6 sin π(20t – 0.5x + π/3)
Therefore, the equation of the resultant wave is:
y = 6 sin π(20t – 0.5x) + 6 sin π(20t – 0.5x + π/3)