Tag: intensity and loudness

Questions Related to intensity and loudness

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

intensity and loudness waves physics

  1. $100$

  2. $200$

  3. $10000$

  4. $10$

Show answer
Correct Option: A
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 $100$

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

intensity and loudness waves physics

  1. $1000$

  2. $2$

  3. $100$

  4. $4$

Show answer
Correct Option: A
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.

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

intensity and loudness waves physics

  1. $2.5$

  2. $ 5 $

  3. $ 50$

  4. $ 10^{5}$

Show answer
Correct Option: D
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 
$\dfrac{I _1}{I _2}=10^5$

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

intensity and loudness waves physics

  1. $5$

  2. $50$

  3. $10^{5}$

  4. $500$

Show answer
Correct Option: C
Explanation:

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

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?

intensity and loudness waves physics

  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

Show answer
Correct Option: B

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 )

intensity and loudness waves physics

  1. $50 dB$

  2. $76 dB$

  3. $68 dB$

  4. $48 dB$

Show answer
Correct Option: C
Explanation:

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

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

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

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

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

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

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? 

intensity and loudness waves physics

  1. 1000

  2. 100

  3. 10,000

  4. None of these

Show answer
Correct Option: A
Explanation:

The intensity of sound in decibel is given by: 


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

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

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

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

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]

intensity and loudness waves physics

  1. $\displaystyle\frac{9r}{10}$

  2. $\displaystyle\frac{r}{10}$

  3. $\displaystyle\frac{3r}{5}$

  4. $\displaystyle\frac{4r}{5}$

Show answer
Correct Option: A
Explanation:

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

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

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

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

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

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