
A point source emits sound equally in all directions in a non absorbing medium. Two points P and Q are at distances of 9m and 25m respectively from the source. The ratio of the amplitude of the waves at P and Q is?
A. $5:3$
B. $3:5$
C. $25:9$
D. $625: 81$
Answer
548.1k+ views
Hint: This is a non absorbing medium, so there will be no loss of energy of the wave as it propagates through the medium. Also, the sound is directed in all directions equally so we do not have to consider the direction. Here we need to find out the ratio of the amplitude and not the intensities.
Complete step by step answer:
We know for a wave propagating through a non-absorbing medium, the amplitude at any point from the source is given by the relation: \[A\alpha \dfrac{1}{x}\], where A is the amplitude and x is the distance from the source to the point where the amplitude is to be determined.
Given \[{{x}_{1}}\]= 9m and \[{{x}_{2}}\]= 25 m
Putting the values, we get,
$A\alpha \dfrac{1}{x} \\
\Rightarrow \dfrac{{{A}_{1}}}{{{A}_{2}}}=\dfrac{{{x}_{2}}}{{{x}_{1}}} \\
\therefore \dfrac{{{A}_{1}}}{{{A}_{2}}}=\dfrac{25}{9}$
So, the correct option is C.
Additional information:
The amplitude of a sound wave is the measure of the height of the wave. The amplitude of a sound wave can be defined as the loudness or the amount of maximum displacement of vibrating particles of the medium from their mean position when the sound is produced. It is the distance between crest or trough and the mean position of the wave.
Note: The amplitude of a wave is related to the amount of energy it carries. A high amplitude wave carries a large amount of energy; a low amplitude wave carries a small amount of energy. The average amount of energy passing through a unit area per unit of time in a specified direction is called the intensity of the wave. As the amplitude of the sound wave increases, the intensity of the sound increases.
Complete step by step answer:
We know for a wave propagating through a non-absorbing medium, the amplitude at any point from the source is given by the relation: \[A\alpha \dfrac{1}{x}\], where A is the amplitude and x is the distance from the source to the point where the amplitude is to be determined.
Given \[{{x}_{1}}\]= 9m and \[{{x}_{2}}\]= 25 m
Putting the values, we get,
$A\alpha \dfrac{1}{x} \\
\Rightarrow \dfrac{{{A}_{1}}}{{{A}_{2}}}=\dfrac{{{x}_{2}}}{{{x}_{1}}} \\
\therefore \dfrac{{{A}_{1}}}{{{A}_{2}}}=\dfrac{25}{9}$
So, the correct option is C.
Additional information:
The amplitude of a sound wave is the measure of the height of the wave. The amplitude of a sound wave can be defined as the loudness or the amount of maximum displacement of vibrating particles of the medium from their mean position when the sound is produced. It is the distance between crest or trough and the mean position of the wave.
Note: The amplitude of a wave is related to the amount of energy it carries. A high amplitude wave carries a large amount of energy; a low amplitude wave carries a small amount of energy. The average amount of energy passing through a unit area per unit of time in a specified direction is called the intensity of the wave. As the amplitude of the sound wave increases, the intensity of the sound increases.
Recently Updated Pages
Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

Which animal has three hearts class 11 biology CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

