Tag: intensity and loudness

Questions Related to intensity and loudness

Which of the following statements are incorrect?

intensity and loudness waves physics

  1. Wave pulses in string are transverse waves

  2. Sound waves in a air are transverse waves of compression and rarefaction

  3. The speed of sound in air at $20^{o}C$ is twice that at $5^{o}C$

  4. A 60 dB sound has twice the intensity of a 30 dB sound

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Correct Option: B,C,D
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?

intensity and loudness waves physics

  1. 10,000

  2. 1000

  3. 100

  4. 10

Show answer
Correct Option: B
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

intensity and loudness waves physics

  1. Intensity level = $\displaystyle\frac{I}{I _0}$ decibels

  2. Intensity level = $I\times I _0$ decibels

  3. Intensity level = $\displaystyle\frac{I}{I _0}\times 100$ decibels

  4. Intensity level = $10\log _{10}\left(\displaystyle\frac{I}{I _0}\right)$ decibels

Show answer
Correct Option: D
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 :

intensity and loudness waves physics

  1. $1\mbox{%}$

  2. $3.01\mbox{%}$

  3. $26\mbox{%}$

  4. $0.1\mbox{%}$

Show answer
Correct Option: C
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 :

intensity and loudness waves physics

  1. $\displaystyle I = Klog L$

  2. $\displaystyle L = K log I$

  3. $\displaystyle L = I$

  4. None of these

Show answer
Correct Option: B
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.

The intensity level due to two waves of the same frequency in a given medium are 1 dB and 4 dB. The ratio of their amplitude is

intensity and loudness waves physics

  1. $1 : 10^4$

  2. $1 : 4$

  3. $1 : 2$

  4. $1 : 10^{0.15}$

Show answer
Correct Option: D
Explanation:

Loudness of the sound       $L = 10  log _{10} \dfrac{I}{I _o}$


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

As  intensity of the wave is directly proportional to the square of the 
amplitude,    i.e       $I   \propto    A^2$

$\implies  $ $L _2 -L _1 = 20  log _{10} \dfrac{A _2}{A _1}$

$4 -  1 = 20  \log _{10} \dfrac{A _2}{A _1}$

Thus   $\dfrac{A _2}{A _1} =   10^{0.15}$

Most of the human ears cannot hear sound of intensity less than :

intensity and loudness waves physics

  1. $\displaystyle 10^{-12} Wm^{-2}$

  2. $\displaystyle 10^{-6} Wm^{-2}$

  3. $\displaystyle 10^{-3} Wm^{-2}$

  4. $\displaystyle 1 Wm^{-2}$

Show answer
Correct Option: A
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?

intensity and loudness waves physics

  1. 13 dB

  2. 7 dB

  3. 4 dB

  4. 2 dB

Show answer
Correct Option: A
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        $             $(log _{10} 20 = 1.3 )$ 

$\implies  L _2 - L _1 = 13    dB$

At what sound level headache begins?

intensity and loudness waves physics

  1. 120-125 dB

  2. 100-105 dB

  3. 80-85 dB

  4. 60-65 dB

Show answer
Correct Option: C
Explanation:

headache starts at 80 dB sound for a normal person.

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. A sound of 60 decibel is ......... times more intense than a sound of 40 decibel :

intensity and loudness waves physics

  1. 20

  2. 100

  3. $\displaystyle 10^{20}$

  4. none of these

Show answer
Correct Option: B
Explanation:

Loudness of two sounds are given as  $L _2 = 60 \ dB$ and $L _1 = 40 \ dB$

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  $60 - 40  = 10\log _{10}\dfrac{I _2}{I _1}$

Or  $\log _{10}\dfrac{I _2}{I _1} = 2$
Or  $\dfrac{I _2}{I _1} = 10^2 = 100$
Thus $60dB$ sound is $100$ times more intense than $40dB$ sound.