Tag: waves

Questions Related to waves

If the intensity of sound is increased by a factor of 30 , by how many decibels is the sound level increased ?

intensity and loudness waves physics

  1. 12 dB

  2. 14.77 dB

  3. 10 dB

  4. 13 dB

Show answer
Correct Option: B
Explanation:
The intensity of sound in increased by factor $30$
Let $2$ be the intensity of previous sound
Intensity of recent sound $I=30I$
$\beta =10 \log\dfrac{I}{I _o}$
$\beta _1=10 \log\dfrac{I}{I _o}$ where intensity of sound is $ I$
$\beta _1=10 \log\dfrac{30I}{I _o}$ when intensity of sound is $I$
Increased in sound level $\Rightarrow \beta _2-\beta _1$
$=10 \log\dfrac{30I}{I _o}-10\log \dfrac{I}{I _o}$
$=10 \log 30$
$10 \log \dfrac{30 I}{I}$
$=14.77d\beta$.

When a person wears a hearing aid, the sound intensity level increases by 30 dB. The sound intensity increases by  

intensity and loudness waves physics

  1. e$^{3}$

  2. 10$^{3}$

  3. 30

  4. 10$^{2}$

Show answer
Correct Option: B
Explanation:

Let the intensity of the sound without hearing Aid is $I _0$, and the 


intensity after wearing hearing aid is $I$, then from the formula, 

$L _{db}=10 \log _{10}{\dfrac{I}{I _0}}$

$30=10 \log _{10}{\dfrac{I}{I _0}}$

$\dfrac{I}{I _0}=10^3$

$I=10^3 \times I _0$

Option "B" is correct.

Spherical sound waves are emitted uniformly in all directions from a point source. The variation in sound level SL as a function of distance 'r' from the source can be written as 

intensity and loudness waves physics

  1. SL = -b log r$^{a}$

  2. SL = a - b (log r)$^{2}$

  3. SL = a - b log r

  4. SL = a - b/r$^{2}$

Show answer
Correct Option: C
Explanation:

we know $SL = 10 log \dfrac{I}{I _o}$


$I = \dfrac{P}{4\pi r^2}$  where P is the power of the source

$SL = 10  log\dfrac{P}{4\pi r^2I _o} = 10  log \dfrac{P}{4\pi I _o} - 10  log  r^2 = 10  log \dfrac{P}{4\pi I _o} - 20  log  r = a - b  log  r$

The intensity level of two sounds are 100 dB and 50 dB. What is the ratio of their intensities?

intensity and loudness waves physics

  1. 10$^{1}$

  2. 10$^{3}$

  3. 10$^{5}$

  4. 10$^{10}$

Show answer
Correct Option: C
Explanation:

Loudness of two sounds is given as  $L _2 = 100 \ dB$ and $L _1 = 50 \ dB$

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  $100 - 50   = 10\log _{10}\dfrac{I _2}{I _1}$

Or  $\log _{10}\dfrac{I _2}{I _1} = 5$
Or  $\dfrac{I _2}{I _1} = 10^5 $

A source of sound emits 200 W power which is uniformly distributed over a sphere of radius 10 m. What is the loudness of sound on the surface of sphere? 

intensity and loudness waves physics

  1. 70dB

  2. 107dB

  3. 80dB

  4. 112dB

Show answer
Correct Option: D
Explanation:

Intensity is given by:
$I=\dfrac{W}{4\pi{r}^2}$
$I=\dfrac{200}{4\pi\times 100} =0.159W/m^2$

In terms of decibels, $I=10\log _{ 10 }{ I } +120 dB$
$I=112dB$

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

Show answer
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.