
The intensity of a sound wave in an elastic medium falls by $10%$ on travelling a distance of 1 m. If the initial intensity of the wave is $100%$ then on travelling a distance of $3 m$ in that medium the intensity will become
A. $81%$
B. $70%$
C. $72.9 %$
D. $60%$
Answer
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Hint: When we consider a sound wave with properties like frequency, amplitude etc. are given to us, we can find quantities like velocity from the formula of bulk modulus. Further the intensity can be derived using these quantities mathematically. Here, we can find a relative relation between two cases and solve the problem.
Complete step by step answer:
We are given that the intensity of a sound wave in an elastic medium falls by $10%$ on travelling a distance of $1 m$ and the initial intensity of the wave is $100%$ .We have to find the intensity of the wave when it further travels a distance of $3m$.Let us take $I$ and $I′$ as the initial intensity and the intensity after traveling a distance of $3m$. As we’re given that the intensity reduces by 10 % when it travels a distance of 1m each,
The intensity of sound after 1m can be written as
$I\prime \prime =90~%\text{ }of\text{ }~I=0.9I~$
Hence after travelling a distance of 3m,
Intensity $I\prime =\left( 0.9 \right)\left[ \left( 0.9 \right)\times 0.9I \right]$
$I\prime =0.729I\\
\therefore I\prime =72.9~%\text{ }of~I$
Hence, option C is the correct answer.
Additional information:
A sound is a vibration that propagates through a medium in the form of a mechanical wave. The medium in which it propagates can either be a solid, a liquid or a gas. Sound travels fastest in solids, relatively slower in liquids and slowest in gases.A sound wave is the pattern of disturbance caused by the energy traveling away from the source of the sound. Sound waves are longitudinal waves. This means that the propagation of vibration of particles is parallel to the direction of the energy wave propagation.
Note:We can mathematically find the intensity of the sound wave using the formula $I=2{{\pi }^{2}}{{f}^{2}}{{\delta }^{2}}\rho v$, where $I$ is the intensity of the sound wave, $f$ is the frequency of the sound, $\delta $ is the amplitude, $\rho $ is the density and v is the velocity of the sound in the medium.
Complete step by step answer:
We are given that the intensity of a sound wave in an elastic medium falls by $10%$ on travelling a distance of $1 m$ and the initial intensity of the wave is $100%$ .We have to find the intensity of the wave when it further travels a distance of $3m$.Let us take $I$ and $I′$ as the initial intensity and the intensity after traveling a distance of $3m$. As we’re given that the intensity reduces by 10 % when it travels a distance of 1m each,
The intensity of sound after 1m can be written as
$I\prime \prime =90~%\text{ }of\text{ }~I=0.9I~$
Hence after travelling a distance of 3m,
Intensity $I\prime =\left( 0.9 \right)\left[ \left( 0.9 \right)\times 0.9I \right]$
$I\prime =0.729I\\
\therefore I\prime =72.9~%\text{ }of~I$
Hence, option C is the correct answer.
Additional information:
A sound is a vibration that propagates through a medium in the form of a mechanical wave. The medium in which it propagates can either be a solid, a liquid or a gas. Sound travels fastest in solids, relatively slower in liquids and slowest in gases.A sound wave is the pattern of disturbance caused by the energy traveling away from the source of the sound. Sound waves are longitudinal waves. This means that the propagation of vibration of particles is parallel to the direction of the energy wave propagation.
Note:We can mathematically find the intensity of the sound wave using the formula $I=2{{\pi }^{2}}{{f}^{2}}{{\delta }^{2}}\rho v$, where $I$ is the intensity of the sound wave, $f$ is the frequency of the sound, $\delta $ is the amplitude, $\rho $ is the density and v is the velocity of the sound in the medium.
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