Two waves of same frequency and intensity ${I_0}$ and $9{I_0}$ produce interference. If at a certain point the resultant intensity is $7{I_0}$ then the minimum phase difference between the two sound waves will be:
A) ${90^ \circ }$
B) ${150^ \circ }$
C) ${120^ \circ }$
D) ${100^ \circ }$
Answer
269.7k+ views
Hint: An objective measure of a wave's time-averaged power density at a given spot. We know the value of two frequencies and the corresponding intensity, so we use the intensity formula to find the difference between two waves when the amplitude of a sound wave is determined by the maximum change in the medium density.
Formula used:
Intensity formula,
$I = {I_1} + {I_2} + 2\sqrt {{I_1}{I_2}} \cos \phi $
Where,
$I$ is the resultant intensity point
${I_1}{I_2}$ are the two waves point
$\cos \phi $ is an amplitude wave angle
Complete step by step solution:
Given by, Let
Intensity wave one \[{I_1} = {I_0}\] , intensity wave second \[{I_2} = 9{I_0}\]
Resultant intensity point $I = 7{I_0}$
According to that the intensity formula,
\[I = {I_1} + {I_2} + 2\sqrt {{I_1}{I_2}} \cos \phi \]
Now we substituting the given value in a above equation
We get,
$\Rightarrow$ \[7{I_0} = {I_0} + 9{I_0} + 2\sqrt {{I_0}\,.9{I_0}} \cos \phi \]
On simplifying, We get,
$\Rightarrow$ \[7{I_0} = 10{I_0} + 2 \times 3{I_0}\cos \phi \]
Therefore, the value $\sqrt 9 $ is $3$
Rearranging the above equation is given below,
$\Rightarrow$ \[7{I_0} - 10{I_0} = 6{I_0}\cos \phi \]
Simplified a given equation,
Here, \[ - 3{I_0} = 6{I_0}\cos \phi \]
Again, we rearranging the given equation
We get,
$\Rightarrow$ \[\cos \phi = - \dfrac{1}{2}\]
According to the trigonometric table
We know that,
Value of $\phi$ is ${120^ \circ }$
then the minimum phase difference between the two sound waves will be ${120^ \circ }$.
Hence, the option C is the correct answer.
Note: As the number of waves passing a reference point is calculated in one second. And the intensity is related to the wave amplitude and the amplitude is squared. The energy of the wave originates from the simple harmonic motion of its particles. The maximum kinetic energy would equal the total energy.
Formula used:
Intensity formula,
$I = {I_1} + {I_2} + 2\sqrt {{I_1}{I_2}} \cos \phi $
Where,
$I$ is the resultant intensity point
${I_1}{I_2}$ are the two waves point
$\cos \phi $ is an amplitude wave angle
Complete step by step solution:
Given by, Let
Intensity wave one \[{I_1} = {I_0}\] , intensity wave second \[{I_2} = 9{I_0}\]
Resultant intensity point $I = 7{I_0}$
According to that the intensity formula,
\[I = {I_1} + {I_2} + 2\sqrt {{I_1}{I_2}} \cos \phi \]
Now we substituting the given value in a above equation
We get,
$\Rightarrow$ \[7{I_0} = {I_0} + 9{I_0} + 2\sqrt {{I_0}\,.9{I_0}} \cos \phi \]
On simplifying, We get,
$\Rightarrow$ \[7{I_0} = 10{I_0} + 2 \times 3{I_0}\cos \phi \]
Therefore, the value $\sqrt 9 $ is $3$
Rearranging the above equation is given below,
$\Rightarrow$ \[7{I_0} - 10{I_0} = 6{I_0}\cos \phi \]
Simplified a given equation,
Here, \[ - 3{I_0} = 6{I_0}\cos \phi \]
Again, we rearranging the given equation
We get,
$\Rightarrow$ \[\cos \phi = - \dfrac{1}{2}\]
According to the trigonometric table
We know that,
Value of $\phi$ is ${120^ \circ }$
then the minimum phase difference between the two sound waves will be ${120^ \circ }$.
Hence, the option C is the correct answer.
Note: As the number of waves passing a reference point is calculated in one second. And the intensity is related to the wave amplitude and the amplitude is squared. The energy of the wave originates from the simple harmonic motion of its particles. The maximum kinetic energy would equal the total energy.
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