Tag: intensity and loudness

Questions Related to intensity and loudness

The power , or loudness, of sound is measured on a scale of increased by a factor of 100$(=10^2$), the sound is called twice as loud and when it increases 10,000 $(=10^4)$ times it is called four times as loud. The exponent of ten is called a bel and one decible is one tenth of a bel. Zero decible is chosen as the intensity of the slowest sound which is just audible or is on the threshold of hearing whereas the intensity of loudest sound is about 170 decibel. Sound intensities above a level of nearly descibels produce a feeling of pain :

  1. 60

  2. 120

  3. 140

  4. 170


Correct Option: B
Explanation:

A human ear can hear a sound of maximum loudness of $120 \ dB$. Sound intensities above nearly $120 \ dB$ produce pain in the human ear.

The intensity of ordinary conversation is rated as nearly 

  1. 10 decibel

  2. 50 decibel

  3. 40 decibel

  4. 60 decibel


Correct Option: D
Explanation:

Human ear can hear sound from 0dB to 140dB. Intensity of ordinary conversation is rated as nearly 60dB.

Two waves of the same pitch have amplitude in the ratio 1 : 3. What will be the ratio of their loudness:

  1. 1:3

  2. 3:1

  3. 1:9

  4. 9:1


Correct Option: C
Explanation:

Loudness depends on the square of the amplitude of the wave. 


That is, $loudness\propto { \left( amplitude \right)  }^{ 2 }$

Thus, when the ratio of amplitudes of the wave is 1:3, the loudness becomes 1:9.

In expressing sound intensity, we take $ 10^{-12} Wm^{-2} $ as the reference level. For ordinary conversion, the intensity level is about $ 10^{-6} Wm^{-2} $. Expressed in decibel, this is :

  1. $ 10^6$

  2. $ 6$

  3. $ 60 $

  4. $ log _e(10^6) $


Correct Option: C
Explanation:

Intensity of reference level  $I _o = 10^{-12} Wm^{-2}$

Measured intensity  $I = 10^{-6} W m^{-2}$
Using  $L = 10\log _{10}\dfrac{I}{I _o}$  decibels
$\therefore  $  $L = 10\log _{10}\dfrac{10^{-6}}{10^{-12}}$   decibels
Or  $L = 10 \ log _{10} 10^{6} = 10\times 6 = 60$  decibels

Decibel (dB) is a unit of loudness of sound. It is defined in a manner such that when amplitude of sound is multiplied by a factor of $\sqrt{10}$, the decibel level increase by 10 units. Loud music of 70 dB is being played at a function. To reduce the loudness to a level of 30 dB, the amplitude of the instrument playing music to be reduced by a factor of

  1. $0$

  2. $10\sqrt{10}$

  3. $100$

  4. $100\sqrt{100}$


Correct Option: C
Explanation:

We know that Loudness of Sound is given by equation,
$\beta  = 10\log \left ( \frac{1}{I _{0}} \right ) dB$
where $I _{0} = 10^{-12} Wm^{-2} $ is the reference unit.
Intensity $ I \propto a^{2} $
where $a$ = amplitude.
Solving we get, the amplitude of the instrument playing music to be reduced by a factor of $100$.

The threshold of hearing for the human ear $ 10^{-12} Wm^{-2} $. This is taken as the standard level. The intensity of sound is $ 1Wm^{-2} $. It has intensity (in $db$)

  1. $10^{12} db $

  2. $12\ db$

  3. $240\ db$

  4. $120\ db$


Correct Option: D
Explanation:

Intensity in $dB$:
$I _{dB} = 10 \log _{10}(\dfrac{I}{I _0})=10 \log _{10}(\dfrac{1}{10^{-12}})=10\times 12 \log _{10}10=120$ $dB$

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


Correct Option: A
Explanation:

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

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

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


Correct Option: D
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.

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

  1. 5

  2. 50

  3. 500

  4. $ 10^5 $


Correct Option: D
Explanation:
Let $I$ represents the intensity of sound.
Loudness of sound          $L = 10  log _{10} \frac{I}{I _o}$
$\implies  L _2 - L _1   = 10   log _{10} \frac{I _2}{I _1}$
$90 - 40  = 10  \log _{10} \dfrac{I _2}{I _1}$
$5= log _{10} \dfrac{I _2}{I _1}              \implies  \dfrac{I _2}{I _1} = 10^5$

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


Correct Option: B
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 $69\,dB$.