
From a line source, if amplitude of a wave at a distance $ r $ is $ A $ , then the amplitude at a distance $ 4r $ will be:-
(A) $ 2A $
(B) $ A $
(C) $ \dfrac{A}{2} $
(D) $ \dfrac{A}{4} $
Answer
571.8k+ views
Hint To solve this question, we have to use the proportionality between the intensity and the amplitude of a wave and the proportionality between the intensity and the distance of the wave. Using these we have to find a relation between amplitude and distance of the wave.
Formula Used: The formula used to solve this question are given as,
$\Rightarrow I \propto {A^2} $
Here, $ I $ is the intensity of the wave and $ A $ is the amplitude of the wave.
$\Rightarrow I \propto \dfrac{1}{{{r^2}}} $
Here, $ r $ is distance travelled by the wave.
Complete step by step answer
We know that Intensity of a wave is inversely proportional to the square of the distance travelled by the wave, i.e.
$\Rightarrow I \propto \dfrac{1}{{{r^2}}} $
Here, $ I $ is the intensity of the wave and $ r $ is the distance travelled by the wave.
Also the intensity can be given as,
$\Rightarrow I = {A^2} $
Here, $ A $ is the amplitude of the wave.
So, the relation between amplitude of the wave and distance travelled by the wave can be given as,
$\Rightarrow {A^2} \propto \dfrac{1}{{{r^2}}} $
Now, if according to the question, the distance travelled by the wave becomes $ 4r $ , then let the corresponding amplitude be $ {A_{new}} $
So, now taking the ratio of the new amplitude to old amplitude we get,
$\Rightarrow \dfrac{{A_{new}^2}}{{{A^2}}} = \dfrac{{\left( {\dfrac{1}{{{{(4r)}^2}}}} \right)}}{{\left( {\dfrac{1}{{{r^2}}}} \right)}} \\
\Rightarrow \dfrac{{A_{new}^2}}{{{A^2}}} = \dfrac{{{r^2}}}{{16{r^2}}} \\
$
Thus, we get the new Amplitude as,
$\Rightarrow
A_{new}^2 = \dfrac{{{A^2}}}{{16}} \\
\Rightarrow {A_{new}} = \dfrac{A}{4} \\
$
$ \therefore $ Option (D) is the correct option out of the given options.
Note
Even though the relations of Intensity with amplitude and distance remain the same in all cases, the relation between distance and amplitude may change depending on the conditions given in the question. So, the latter should always be deduced while solving the question.
Formula Used: The formula used to solve this question are given as,
$\Rightarrow I \propto {A^2} $
Here, $ I $ is the intensity of the wave and $ A $ is the amplitude of the wave.
$\Rightarrow I \propto \dfrac{1}{{{r^2}}} $
Here, $ r $ is distance travelled by the wave.
Complete step by step answer
We know that Intensity of a wave is inversely proportional to the square of the distance travelled by the wave, i.e.
$\Rightarrow I \propto \dfrac{1}{{{r^2}}} $
Here, $ I $ is the intensity of the wave and $ r $ is the distance travelled by the wave.
Also the intensity can be given as,
$\Rightarrow I = {A^2} $
Here, $ A $ is the amplitude of the wave.
So, the relation between amplitude of the wave and distance travelled by the wave can be given as,
$\Rightarrow {A^2} \propto \dfrac{1}{{{r^2}}} $
Now, if according to the question, the distance travelled by the wave becomes $ 4r $ , then let the corresponding amplitude be $ {A_{new}} $
So, now taking the ratio of the new amplitude to old amplitude we get,
$\Rightarrow \dfrac{{A_{new}^2}}{{{A^2}}} = \dfrac{{\left( {\dfrac{1}{{{{(4r)}^2}}}} \right)}}{{\left( {\dfrac{1}{{{r^2}}}} \right)}} \\
\Rightarrow \dfrac{{A_{new}^2}}{{{A^2}}} = \dfrac{{{r^2}}}{{16{r^2}}} \\
$
Thus, we get the new Amplitude as,
$\Rightarrow
A_{new}^2 = \dfrac{{{A^2}}}{{16}} \\
\Rightarrow {A_{new}} = \dfrac{A}{4} \\
$
$ \therefore $ Option (D) is the correct option out of the given options.
Note
Even though the relations of Intensity with amplitude and distance remain the same in all cases, the relation between distance and amplitude may change depending on the conditions given in the question. So, the latter should always be deduced while solving the question.
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