Physics · Science General

Acoustics and Sound Waves

2,006 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

Multiple choice intensity and loudness waves physics

The threshold of sound is $ 10^{-12} Wm^{-2} $.What is the intensity level of sound whose intensity is $10^{-8} Wm^{-2} $?

  1. 40 db

  2. 8db

  3. 12 db

  4. 20 db
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Intensity in decibels is expressed as $I(dB)=10log(\dfrac{I}{I _0})$

where $I _0$ is the threshold intensity of sound.
Thus $I(dB)=10log(\dfrac{10^{-8}}{10^{-12}})=40$

Multiple choice intensity and loudness waves physics

A key of a mechanical piano is struck gently and then struck again but much harder this time. In the second case

  1. both loudness and pitch will remain unaffected

  2. sound will be louder and pitch will also be higher

  3. sound will be louder but pitch will be lower

  4. sound will be louder but pitch will not be different
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

When the key of a mechanical piano struck harder, the vibrations caused are more resulting in larger amplitude of the wave that propagates from the key.
Loudness depends on the square of the amplitude of the wave. 
That is, $loudness\propto { \left( amplitude \right)  }^{ 2 }$
Amplitude is the size of the vibration, and this determines how loud the sound is.  Larger vibrations make a louder sound.
Pitch of the sound depends on the frequency of the wave which is not change when the key is struck harder.

Multiple choice intensity and loudness waves physics

How many times more intense is 90 dB sound than 40 dB sound :

  1. 5

  2. 50

  3. 500
  4. $ 10^5 $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Let $I$ represents the intensity of sound.
Loudness of sound          $L = 10  log _{10} \frac{I}{I _o}$
$\implies  L _2 - L _1   = 10   log _{10} \frac{I _2}{I _1}$
$90 - 40  = 10  \log _{10} \dfrac{I _2}{I _1}$
$5= log _{10} \dfrac{I _2}{I _1}              \implies  \dfrac{I _2}{I _1} = 10^5$
Multiple choice intensity and loudness waves physics

A person is talking in a small room and the sound intensity level is 60 dB everywhere in the room. If there are eight people talking simultaneously in the room, what is the sound intensity level?

  1. 60 dB

  2. 69 dB

  3. 74 dB

  4. 81 dB
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let $I$ be the  intensity due to a person. For 8 persons, the  intensity, $I^{'}=8I$


Given, the sound intensity level, $B _1$ inside the room is 60 dB. We have to find out the sound intensity level when  eight people talking simultaneously in the room. Let it be $B _2$

We have the equation,

$\triangle B= 10 \, log \, (\frac{I^{'}}{I})$

$B _2-B _1=10 \, log \, (\frac{I{'}}{I})$


$B _2-60=10 \, log \, (\frac{8I}{I})$


$B _2-60=10 \, log \,8$

$B _2-60=10 \, log \,2^3$

$B _2=9=69\,dB$

Hence the sound intensity level is $69\,dB$.

Multiple choice intensity and loudness waves physics

A sound absorber attenuates the sound level by $20 dB$. The intensity decreases by a factor of

  1. $100$
  2. $200$
  3. $10000$
  4. $10$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We have, $\displaystyle L _1=10log \left ( \frac {I _1}{I _0}\right)$

$\displaystyle L _2=10log \left ( \frac {I _2}{I _0}\right)$

$\displaystyle \therefore L _1 - L _2=10log \left ( \frac {I _1}{I _0}\right) - 10log \left ( \frac {I _2}{I _0}\right)$

or, $\displaystyle \Delta L = 10log \left ( \frac {I _1}{I _0} \times \frac {I _0}{I _2}\right)$

or, $\displaystyle \Delta L = 10log \left ( \frac {I _1}{I _2}\right)$

or, $\displaystyle 20 = 10log \left ( \frac {I _1}{I _2}\right)$

or, $\displaystyle 2 = log \left ( \frac {I _1}{I _2}\right)$

or, $\displaystyle \frac {I _1}{I _2}=10^2$

or, $\displaystyle I _2 = \frac {I _1}{100}$

$\Rightarrow $ Intensity decreases by a factor $100$

Multiple choice intensity and loudness waves physics

How many times more intense is a $60 \ dB$ sound than a $30 \ dB$ sound?

  1. $1000$
  2. $2$
  3. $100$
  4. $4$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Loudness of two sounds are given as  $L _2 = 60 \ dB$ and $L _1 = 30 \ dB$

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  $60 - 30  = 10\log _{10}\dfrac{I _2}{I _1}$

Or  $\log _{10}\dfrac{I _2}{I _1} = 3$
Or  $\dfrac{I _2}{I _1} = 10^3 = 1000$
Thus $60dB$ sound is $1000$ times more intense than $30dB$ sound.

Multiple choice intensity and loudness waves physics

How many times more intense is a $ 90\, dB$ sound than a $ 40\, dB$ sound ?

  1. $2.5$
  2. $ 5 $
  3. $ 50$
  4. $ 10^{5}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$90=10 log _{10}(\dfrac{I _1}{I _0})$
$\Rightarrow    \dfrac{I _1}{I _0}=10^9$......(i)
Again, $40=10 log _{10}(\dfrac{I _2}{I _0})$
$\Rightarrow    \dfrac{I _2}{I _0}=10^4$......(ii)
From Eqs.(i) and (ii), we get 
$\dfrac{I _1}{I _2}=10^5$

Multiple choice intensity and loudness waves physics

$90dB$ sound is $'x'$ times more intense than $40dB$ sound, then $x$ is

  1. $5$
  2. $50$
  3. $10^{5}$
  4. $500$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For every 10dB, intensity rises by 10 times, so for 50 dB intensity will rise by $10^5$

Multiple choice intensity and loudness waves physics

A hearing test is conducted on an aged person. It is found that her threshold of hearing is $20$ decibels at $1$ kHz and it rises linearly with frequency to $60$ decibels at $9$ kHz. The minimum intensity of sound that the person can hear at $5$ kHz is?

  1. $10$ times than that at $1$ kHz
  2. $100$ times than that at $1$ kHz
  3. $0.5$ times than that at $9$ kHz
  4. $0.05$ times than that at $9$ kHz
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The threshold rises from 20 dB at 1 kHz to 60 dB at 9 kHz. The slope is (60-20)/(9-1) = 5 dB/kHz. At 5 kHz, the threshold is 20 + (5 * 4) = 40 dB. The difference between 40 dB and 20 dB is 20 dB, which corresponds to a factor of 10^(20/10) = 100.

Multiple choice intensity and loudness waves physics

A dog while barking delivers about $1 mW$  of power. If this power is uniformly distributed over a hemispherical area, the sound level at a distance of $5 m$ is (given  10 log$ _{10}$ 6.37 $=$0.8 )

  1. $50 dB$
  2. $76 dB$
  3. $68 dB$
  4. $48 dB$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$I = \dfrac{Power}{Surface Area}$

$I = 6.37 \times 10^{-6} Wm^{-2}$

$I _{o} = 10^{-12} Wm^{-2}$

$SL = 10 log (\dfrac{I}{I _{o}})$

$SL = 10 log (\dfrac{6.37 \times 10^{-6}}{10^{-12}})$

$SL = 10 log (6.37) + 10 log(10^{6})SL = 8 + 60 = 68\ dB$

Multiple choice intensity and loudness waves physics

When a sound wave enters the ear, it sets the eardrum into oscillation, which  in turn causes oscillation of 3 tiny bones in the middle ear called ossicles. This oscillation is finally transmitted to the fluid filled in inner portion of the ear termed as inner ear, the motion of the fluid disturbs hair calls within the inner ear which transmit nerve impulses to the brain with information that a sound is present. The three bones present in the middle ear are named as hammer, anvil and stirrup. Out of these the stirrup is the smallest one and this only connects the middle  ear to inner ear as shown in the figure below. The area of stirrup and its extent of connection with the inner ear limits the sensitivity of the human ear. Consider a person's eat whose moving part of the eardrum has an area of about 43 mm$^{2}$ and the area of stirrup is about 3.2 mm$^{2}$. The mass of ossicles is negligible. As a result, force  exerted by sound wave in air on eardrum and ossicles is same as the force exerted by ossicles on the inner ear. Consider a sound wave having maximum pressure fluctuation of $3\times10^{-2}$ Pa from its normal equilibrium pressure value which is wqual to $10^{5}$ Pa. Frequency of sound wave is 1200 Hz. 
Data: Velocity of sound wave in air is  332 m/s. Velocity of sound wave in fluid (present in inner ear) is 1500 m/s. Bulk modulus of air is $1.42\times10^{5}$ Pa. Bulk modulus of fluid is $2.18\times10^{9}$ Pa. 


If then person is using an hearing aid, which increase the sound intensity level by 30 dB, then by what factor the intensity of given sound wave change as perceived by inner ear? 

  1. 1000

  2. 100

  3. 10,000

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The intensity of sound in decibel is given by: 


$I _{dB}=10log(\dfrac{I}{I _0})$

Hence, $30=10log(\dfrac{I}{I _0})$

$\implies I=10^3I _0=1000I _0$

$\implies \dfrac{I}{I _0}=1000$

Multiple choice intensity and loudness waves physics

A bird is singing on a tree and a man is hearing at a distance $'r'$ from the bird. Calculate the displacement of the man towards the bird so that the loudness heard by man increases by $20\;dB$.
[Assume that the motion of man is along the line joining the bird and the man]

  1. $\displaystyle\frac{9r}{10}$
  2. $\displaystyle\frac{r}{10}$
  3. $\displaystyle\frac{3r}{5}$
  4. $\displaystyle\frac{4r}{5}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Loudness $\beta=10\;log _{10}\displaystyle\frac{I}{I _0}$

$\therefore\;\beta _2-\beta _1=10\;log _{10}\displaystyle\frac{I _2}{I _1}$

$\because\;I=\displaystyle\frac{P}{4\pi r^2}\;\therefore\;\displaystyle\frac{I _2}{I _1}=\displaystyle\frac{r _1^2}{r _2^2}$

$\therefore\;(\beta+20)-\beta=10\;log _{10}\displaystyle\frac{r^2}{r _2^2}=20\;log _{10}\displaystyle\frac{r}{r _2}$

$\Rightarrow \displaystyle\frac{r}{r _2}=10\Rightarrow r _2=0.1\;r$

$\therefore\;shift=r-0.1\;r=0.9\;r$.

Multiple choice biology sensory organs of humans - sense organs ear : the organ of hearing and equilibrium ear and its function ear

Which of the following membranes given us the ability to different pitches of sound?

  1. Membrane covering the round window

  2. Tympanic membrane

  3. Tectorial membrane

  4. Basilar membrane
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The human ear can detect a wide range of frequencies, from the low rumbles of distant thunder to the high-pitched whine of a mosquito. The sensory cells that detect these sounds are called hair cells, named for the hair-like strands that cluster on their tops. Hair cells are spread across a flat surface called the basilar membrane, which is rolled like a carpet and tucked into a snail shell-shaped structure in the inner ear called the cochlea. Each of our roughly 16,000 hair cells is dedicated to a narrow frequency range. These cells are ordered along the basilar membrane according to the frequencies they detect.

So the correct option is 'Basilar membrane'.