Physics · Science General
Acoustics and Sound Waves
2,160 Questions
Acoustics and sound waves deal with mechanical vibrations traveling through media like air and water. Key concepts include wave reflection, beats, echoes, the Mach number, and infrasound. These physics fundamentals are regularly tested in general science sections of multiple competitive exams.
Sound wave propagationEchoes and reflectionWave interferenceMach numberInfrasound frequency
Acoustics and Sound Waves Questions
Identify the scale which can be used to measure the intensity of sound?
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decibel
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acoustic
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ultrasound
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infrasound
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Hertz
A
Correct answer
Explanation
Intensity (or loudness) of sound $L =10 \log _{10} \dfrac{I}{I _o}$
It is measured in decibel (dB)
A normal human being can hear sound having an intensity level of maximum ...................
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50 dB
-
80 dB
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100 dB
-
150 dB
B
Correct answer
Explanation
Decibel levels help us to understand the loudness of a sound with respect to a reference level. A decibel level of zero dB represents the lowest level of sound and a level of 80 dB represents the level of pain for humans.
How many order of magnitude more powerful is 90 dB sound than 40 dB sound?
D
Correct answer
Explanation
$90 = 10 log (\dfrac{I _{1}}{I})$
$40 = 10 log (\dfrac{I _{2}}{I})$
$90 - 40 = 10 log (\dfrac{I _{1}}{I _{2}})$
$\dfrac{I _{1}}{I _{2}} = 10^{5}$
If the intensity of sound is increased by a factor of 30 , by how many decibels is the sound level increased ?
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12 dB
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14.77 dB
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10 dB
-
13 dB
B
Correct answer
Explanation
The intensity of sound in increased by factor $30$
Let $2$ be the intensity of previous sound
Intensity of recent sound $I=30I$
$\beta =10 \log\dfrac{I}{I _o}$
$\beta _1=10 \log\dfrac{I}{I _o}$ where intensity of sound is $ I$
$\beta _1=10 \log\dfrac{30I}{I _o}$ when intensity of sound is $I$
Increased in sound level $\Rightarrow \beta _2-\beta _1$
$=10 \log\dfrac{30I}{I _o}-10\log \dfrac{I}{I _o}$
$=10 \log 30$
$10 \log \dfrac{30 I}{I}$
$=14.77d\beta$.
When a person wears a hearing aid, the sound intensity level increases by 30 dB. The sound intensity increases by
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e$^{3}$
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10$^{3}$
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30
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10$^{2}$
B
Correct answer
Explanation
Let the intensity of the sound without hearing Aid is $I _0$, and the
intensity after wearing hearing aid is $I$, then from the formula,
$L _{db}=10 \log _{10}{\dfrac{I}{I _0}}$
$30=10 \log _{10}{\dfrac{I}{I _0}}$
$\dfrac{I}{I _0}=10^3$
$I=10^3 \times I _0$
Option "B" is correct.
Spherical sound waves are emitted uniformly in all directions from a point source. The variation in sound level SL as a function of distance 'r' from the source can be written as
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SL = -b log r$^{a}$
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SL = a - b (log r)$^{2}$
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SL = a - b log r
-
SL = a - b/r$^{2}$
C
Correct answer
Explanation
we know $SL = 10 log \dfrac{I}{I _o}$
$I = \dfrac{P}{4\pi r^2}$ where P is the power of the source
$SL = 10 log\dfrac{P}{4\pi r^2I _o} = 10 log \dfrac{P}{4\pi I _o} - 10 log r^2 = 10 log \dfrac{P}{4\pi I _o} - 20 log r = a - b log r$
The intensity level of two sounds are 100 dB and 50 dB. What is the ratio of their intensities?
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10$^{1}$
-
10$^{3}$
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10$^{5}$
-
10$^{10}$
C
Correct answer
Explanation
Loudness of two sounds is given as $L _2 = 100 \ dB$ and $L _1 = 50 \ dB$
Loudness of sound $L = 10 \log _{10}\dfrac{I}{I _o}$
$\implies$ $L _2 - L _1 = 10\log _{10}\dfrac{I _2}{I _1}$
Or $100 - 50 = 10\log _{10}\dfrac{I _2}{I _1}$
Or $\log _{10}\dfrac{I _2}{I _1} = 5$
Or $\dfrac{I _2}{I _1} = 10^5 $
A source of sound emits 200 W power which is uniformly distributed over a sphere of radius 10 m. What is the loudness of sound on the surface of sphere?
D
Correct answer
Explanation
Intensity is given by:
$I=\dfrac{W}{4\pi{r}^2}$
$I=\dfrac{200}{4\pi\times 100} =0.159W/m^2$
In terms of decibels, $I=10\log _{ 10 }{ I } +120 dB$
$I=112dB$
Which of the following statements are incorrect?
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Wave pulses in string are transverse waves
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Sound waves in a air are transverse waves of compression and rarefaction
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The speed of sound in air at $20^{o}C$ is twice that at $5^{o}C$
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A 60 dB sound has twice the intensity of a 30 dB sound
B,C,D
Correct answer
Explanation
Case B : Sound wave in air is the longitudinal wave. so B is not true
Case C because of higher temperature means molecules having higher energy and moving faster than in cold temperature
case D :
$L = 10 log \dfrac{I}{I _o}$
$\dfrac{60}{30} = \dfrac{10 log\dfrac{I _1}{I _o}}{10 log\dfrac{I _2}{I _o}}$
$2log\dfrac{I _2}{I _o} = log \dfrac{I _1}{I _o}$
so $I _2 = 2I _1$ is not true.
If the loudness changes from 30 dB to 60 dB. What is the ratio of the intensities in two cases?
B
Correct answer
Explanation
Let the intensities be $I _1$ and $I _2$ for loudness $L _1 = 30 dB$ and $L _2 = 60 dB$ respectively.
Loudness of sound $L = 10 log _{10} \dfrac{I}{I _o}$ where $I$ is the intensity of the sound wave
$\implies L _2 - L _1 = 10 log _{10} \dfrac{I _2}{I _1}$
Thus $60 - 30 = 10 log _{10} \dfrac{I _2}{I _1}$
$\implies \dfrac{I _2}{I _1} = 10^3$
The intensity level of sound having intensity $I$ is defined in term of $I _0$. The threshold intensity of hearing is
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Intensity level = $\displaystyle\frac{I}{I _0}$ decibels
-
Intensity level = $I\times I _0$ decibels
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Intensity level = $\displaystyle\frac{I}{I _0}\times 100$ decibels
-
Intensity level = $10\log _{10}\left(\displaystyle\frac{I}{I _0}\right)$ decibels
D
Correct answer
Explanation
The intensity level of sound having intensity I is defined in term of $I _0$. The threshold intensity of hearing is $I _{dB} = 10 \log _{10}({\dfrac{I}{I _0}})$.
An increase in the intensity level of one decibel implies an increase in intensity of :
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$1\mbox{%}$
-
$3.01\mbox{%}$
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$26\mbox{%}$
-
$0.1\mbox{%}$
C
Correct answer
Explanation
The Decibel scale is given by
L = $10log(\frac{I}{I{ _{o}}})$ ; $I{ _{o}}$ is the intensity at threshold for hearing.
Let the reading be X;
X = $10log(\frac{I{ _{1}}}{I{ _{0}}})$;
X + 1 = $10log(\frac{I{ _{2}}}{I{ _{0}}})$;
Subtracting the equations we get
$\Rightarrow $1 = 10($log\dfrac{I{ _{2}}}{I{ _{0}}} - log\dfrac{I{ _{1}}}{I{ _{0}}}$)
$\Rightarrow $1 = 10($log\dfrac{I{ _{2}}}{I{ _{1}}}$)
$\Rightarrow $$log\dfrac{I{ _{2}}}{I{ _{1}}}$ = $ \dfrac{1}{10}$
$\Rightarrow $$\dfrac{I{ _{2}}}{I{ _{1}}}$ = $10^{0.1}$ ; $10^{0.1}$ = 1.26;
$\Rightarrow $$\dfrac{I{ _{2}}}{I{ _{1}}}$ = 1.26;
$\Rightarrow $$I{ _{2}}$=1.26$I{ _{1}}$; And hence a 26% increase
The relation between the objective measurement of intensity of sound, $I$ and the subjective sensory response called loudness $L$ is given by :
-
$\displaystyle I = Klog L$
-
$\displaystyle L = K log I$
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$\displaystyle L = I$
-
None of these
B
Correct answer
Explanation
loudness($L$) in terms of intensity ($I$) is
$L=K\log(I)$
Where $L$ is the loudness, $I$ is the intensity and $K$ is a constant of proportionality.
so the answer is B.
Most of the human ears cannot hear sound of intensity less than :
-
$\displaystyle 10^{-12} Wm^{-2}$
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$\displaystyle 10^{-6} Wm^{-2}$
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$\displaystyle 10^{-3} Wm^{-2}$
-
$\displaystyle 1 Wm^{-2}$
A
Correct answer
Explanation
A human ear can hear a sound of minimum intensity $10^{-12} W/m^2$. This intensity sound corresponds to loudness $0 \ dB$. So, we say that a human ear cannot hear a sound of loudness $0 \ dB$.
A human ear can hear a sound of maximum intensity $1 \ W/m^2.$
The power of a loud speaker is increased from 20 W to 400 W. What is the intensity increase as compared to the original value?
A
Correct answer
Explanation
As intensity of wave is directly proportional to power, i.e $I \propto P$
Loudness of sound, $L = 10 log _{10} \dfrac{I}{I _o}$
$\implies L _2 - L _1 = 10 log _{10} \dfrac{I _2}{I _1}$
Thus, $ L _2 - L _1 = 10 log _{10} \dfrac{P _2}{P _1}$
$ L _2 - L _1 = 10 log _{10} \dfrac{400}{20}$
$ L _2 - L _1 = 10 log _{10} 20 $ $(log _{10} 20 = 1.3 )$
$\implies L _2 - L _1 = 13 dB$