Tag: intensity and loudness

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Questions Related to intensity and loudness

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

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

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

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

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

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