
How can I calculate the interference of waves?
Answer
457.5k+ views
Hint: Wave is known as propagating dynamic disturbance which is the change from the equilibrium of one or more quantities, it is also described by a wave equation. Waves can be periodic, those quantities oscillate again and again about an equilibrium value at frequency.
Complete step by step answer:
Two waves of equal amplitude (A) and equal frequency w, where,\[w = 2\pi f\] with a phase difference \[\phi \]
Equation of two waves are
\[
x_1 = Acoswt \\
x_2 = Acos(wt + \phi ) \\
\]:
By using this trigonometric identity
\[cos(x) + cos(y) = 2cos(\dfrac{{x + y}}{2})cos(\dfrac{{x - y}}{2})\]
By adding first and second waves
Then the resultant wave ${x_r}$ is:
\[
{x_r} = x_1 + x_2 \\
{x_r} = Acoswt + Acos(wt + \phi ) \\
\]
By simplifying the above equation,
\[{x_r} = 2Acos(\dfrac{\phi }{2})cos(\dfrac{{wt + \phi }}{2})\]
There is constructive interference or destructive interference.
First for constructive interference,
the above equation will be, \[cos(\dfrac{\phi }{2}) = 1\]
It will give values for \[\phi = 0,2\pi ,4\pi etc\]
For destructive interference,
the above equation is, \[cos(\dfrac{\phi }{2}) = 0\]
It gives values as \[\phi = \pi ,3\pi ,5\pi etc\]
If the frequencies of waves are different, but there is no phase difference:
Then it gives values for two waves
\[
x_1 = Acosw_1t \\
x_2 = Acos(w_2t) \\
\]
By using the identity then the Resultant superposed wave is
\[{x_r} = 2Acos(\dfrac{{w_1 + w_2}}{2})cos(\dfrac{{w_1 - w_2}}{2})\]
The ${x_r}$ is the resultant wave equation. Wave is the product of two waves in which the sum and difference of the original waves, then it will be beats.
Note:
Wave velocity is the velocity with which the wave travels through the medium. The velocity of a sound wave is maximum in solids because they are more elastic in nature than liquids and gases. There are two types of waves. There are sound waves and light waves. Medium is required for the propagation of sound waves where it is longitudinal. Medium is not required for the propagation of light waves which are transverse.
Complete step by step answer:
Two waves of equal amplitude (A) and equal frequency w, where,\[w = 2\pi f\] with a phase difference \[\phi \]
Equation of two waves are
\[
x_1 = Acoswt \\
x_2 = Acos(wt + \phi ) \\
\]:
By using this trigonometric identity
\[cos(x) + cos(y) = 2cos(\dfrac{{x + y}}{2})cos(\dfrac{{x - y}}{2})\]
By adding first and second waves
Then the resultant wave ${x_r}$ is:
\[
{x_r} = x_1 + x_2 \\
{x_r} = Acoswt + Acos(wt + \phi ) \\
\]
By simplifying the above equation,
\[{x_r} = 2Acos(\dfrac{\phi }{2})cos(\dfrac{{wt + \phi }}{2})\]
There is constructive interference or destructive interference.
First for constructive interference,
the above equation will be, \[cos(\dfrac{\phi }{2}) = 1\]
It will give values for \[\phi = 0,2\pi ,4\pi etc\]
For destructive interference,
the above equation is, \[cos(\dfrac{\phi }{2}) = 0\]
It gives values as \[\phi = \pi ,3\pi ,5\pi etc\]
If the frequencies of waves are different, but there is no phase difference:
Then it gives values for two waves
\[
x_1 = Acosw_1t \\
x_2 = Acos(w_2t) \\
\]
By using the identity then the Resultant superposed wave is
\[{x_r} = 2Acos(\dfrac{{w_1 + w_2}}{2})cos(\dfrac{{w_1 - w_2}}{2})\]
The ${x_r}$ is the resultant wave equation. Wave is the product of two waves in which the sum and difference of the original waves, then it will be beats.
Note:
Wave velocity is the velocity with which the wave travels through the medium. The velocity of a sound wave is maximum in solids because they are more elastic in nature than liquids and gases. There are two types of waves. There are sound waves and light waves. Medium is required for the propagation of sound waves where it is longitudinal. Medium is not required for the propagation of light waves which are transverse.
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