Tag: waves

Questions Related to waves

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 $db$)

intensity and loudness waves physics

  1. $10^{12} db $

  2. $12\ db$

  3. $240\ db$

  4. $120\ db$

Show answer
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} $?

intensity and loudness waves physics

  1. 40 db

  2. 8db

  3. 12 db

  4. 20 db

Show answer
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

intensity and loudness waves physics

  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

Show answer
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 :

intensity and loudness waves physics

  1. 5

  2. 50

  3. 500

  4. $ 10^5 $

Show answer
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?

intensity and loudness waves physics

  1. 60 dB

  2. 69 dB

  3. 74 dB

  4. 81 dB

Show answer
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 $69\,dB$.

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