
A person is standing at a distance D from an isotropic point source of sound. He walks $50.0m$ towards the source and observes that the intensity of the sound has doubled. His initial distance D from the source is?
A. $50\sqrt 2 m$
B. $\dfrac{{50\sqrt 2 }}{{\sqrt 2 - 1}}m$
C. $\dfrac{{50}}{{\sqrt 2 - 1}}m$
D. $100\sqrt 2 m$
Answer
232.8k+ views
Hint: In the case of a problem based on wave and acoustics phenomena, we know that the Intensity of the sound varies inversely with distance hence, use the mathematical relation between the two which is given in form of $I = \dfrac{P}{{4\pi {D^2}}}$ to find the solution of the given problem and select a correct option given.
Formula used:
the Intensity of sound is given as:
$I = \dfrac{P}{{4\pi {D^2}}}$
where P = sound power
D = Distance travelled
Complete step by step solution:
As we know that the Intensity of sound is given as:
$I = \dfrac{P}{{4\pi {D^2}}}$
Now, according to the question, there will be 2 cases:
Case1. Initially when a person is standing from the source at a distance D.
In this case, $D = D$ and $I = I$
$I = \dfrac{P}{{4\pi {D^2}}}$ … (1)
Case2. When a person walks $50.0m$ towards the source, the intensity of the sound gets doubled.
In this case, $D = D - 50$ and $I = 2I$( $\therefore $Sound Power will remain same in both the cases)
$2I = \dfrac{P}{{4\pi {{(D - 50)}^2}}}$ … (2)
Divide eq. (1) by eq. (2), we get
$\dfrac{I}{{2I}} = \dfrac{{\dfrac{P}{{4\pi {D^2}}}}}{{\dfrac{P}{{4\pi {{(D - 50)}^2}}}}} = \dfrac{{{{(D - 50)}^2}}}{{{D^2}}}$
$ \Rightarrow \dfrac{1}{2} = \dfrac{{{{(D - 50)}^2}}}{{{D^2}}} = {\left( {\dfrac{{D - 50}}{D}} \right)^2}$
Taking Square roots on both sides, we get
$\dfrac{{D - 50}}{D} = \dfrac{1}{{\sqrt 2 }}$
$\therefore D = \dfrac{{50\sqrt 2 }}{{\sqrt 2 - 1}}m$
Thus, the initial distance D of a person from the source is $\dfrac{{50\sqrt 2 }}{{\sqrt 2 - 1}}m$.
Hence, the correct option is B.
Note: Sound intensity can be measured in energy or work units, such as microjoules per second per square centimetre, or in power units, such as microwatts per square centimetre. Sound intensity, unlike loudness, is objective and can be evaluated by auditory apparatus independent of an observer's hearing.
Formula used:
the Intensity of sound is given as:
$I = \dfrac{P}{{4\pi {D^2}}}$
where P = sound power
D = Distance travelled
Complete step by step solution:
As we know that the Intensity of sound is given as:
$I = \dfrac{P}{{4\pi {D^2}}}$
Now, according to the question, there will be 2 cases:
Case1. Initially when a person is standing from the source at a distance D.
In this case, $D = D$ and $I = I$
$I = \dfrac{P}{{4\pi {D^2}}}$ … (1)
Case2. When a person walks $50.0m$ towards the source, the intensity of the sound gets doubled.
In this case, $D = D - 50$ and $I = 2I$( $\therefore $Sound Power will remain same in both the cases)
$2I = \dfrac{P}{{4\pi {{(D - 50)}^2}}}$ … (2)
Divide eq. (1) by eq. (2), we get
$\dfrac{I}{{2I}} = \dfrac{{\dfrac{P}{{4\pi {D^2}}}}}{{\dfrac{P}{{4\pi {{(D - 50)}^2}}}}} = \dfrac{{{{(D - 50)}^2}}}{{{D^2}}}$
$ \Rightarrow \dfrac{1}{2} = \dfrac{{{{(D - 50)}^2}}}{{{D^2}}} = {\left( {\dfrac{{D - 50}}{D}} \right)^2}$
Taking Square roots on both sides, we get
$\dfrac{{D - 50}}{D} = \dfrac{1}{{\sqrt 2 }}$
$\therefore D = \dfrac{{50\sqrt 2 }}{{\sqrt 2 - 1}}m$
Thus, the initial distance D of a person from the source is $\dfrac{{50\sqrt 2 }}{{\sqrt 2 - 1}}m$.
Hence, the correct option is B.
Note: Sound intensity can be measured in energy or work units, such as microjoules per second per square centimetre, or in power units, such as microwatts per square centimetre. Sound intensity, unlike loudness, is objective and can be evaluated by auditory apparatus independent of an observer's hearing.
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