Sound Intensity and Loudness

The increase in decibels is calculated as 10 * log10(P2/P1). Here, 10 * log10(400/10) = 10 * log10(40) = 10 * 1.602 = 16.02 dB.

Amplitude is the size of the vibration, and this determines how loud the sound is Larger vibrations make a louder sound.The intensity or loudness of a sound depends upon the extent to which the sounding body vibrates, i.e., the amplitude of vibration. A sound is louder as the amplitude of vibration is greater, and the intensity decreases as the distance from the source increases. Hence, loudness is dependent on amplitude.

Amplitude is the size of the vibration, and this determines how loud the sound is.  Larger vibrations make a louder sound.
The intensity or loudness of a sound depends upon the extent to which the sounding body vibrates, i.e., the amplitude of vibration. A sound is louder as the amplitude of vibration is greater, and the intensity decreases as the distance from the source increases.
Hence, loudness of the sound is affected due to the change in its amplitude.

The intensity or loudness of a sound depends upon the extent to which the sounding body vibrates, i.e., the amplitude of vibration. A sound is louder as the amplitude of vibration is greater, and the intensity decreases as the distance from the source increases. Loudness is measured in units called decibels. 
Loudness depends on the square of the amplitude of the wave. That is, $loudness{ \propto \left( amplitude \right)  }^{ 2 }$.
Hence, As it moves away from the source its amplitude as well as its loudness decreases.

The loudness of sound is directly proportional to amplitude of sound wave. 

Unit for measuring the loudness of sound is decibel.
The intensity or loudness of a sound depends upon the extent to which the sounding body vibrates, i.e., the amplitude of vibration. A sound is louder as the amplitude of vibration is greater, and the intensity decreases as the distance from the source increases. Loudness is measured in units called decibels. 

Answer is B.

The intensity or loudness of a sound depends upon the extent to which the sounding body vibrates, i.e., the amplitude of vibration. A sound is louder as the amplitude of vibration is greater, and the intensity decreases as the distance from the source increases.
Sound levels is measured in units called decibels. 

If the amplitude of the sound is made 3 times its initial value, then its loudness will become nine times its initial value.
Loudness depends on the square of the amplitude of the wave. 
That is, $Loudness\propto { (Amplitude) }^{ 2 }$.
Thus, when amplitude of the wave is tripled, the loudness increases nine times.

Answer is C.

The intensity or loudness of a sound depends upon the extent to which the sounding body vibrates, i.e., the amplitude of vibration. A sound is louder as the amplitude of vibration is greater, and the intensity decreases as the distance from the source increases. Loudness or intensity of sound is measured in units called decibels. 

The loudness of sound is measured in units called decibels (dB). A decibel unit expresses the relative intensity of sounds on a scale from zero for the average least perceptible sound to about 100 dB, which is near the level most people find uncomfortably loud.

In S. I. unit of intensity of sound is
joule s$^{-1}$ m$^{-2}$ or watt m$^{-2}$
Intensity = sound energy/time $\times$ area

Intensity (or loudness) of sound         $L =10 \log _{10} \dfrac{I}{I _o}$

Decibel levels help us to understand the loudness of a sound with respect to a reference level. A decibel level of zero dB represents the lowest level of sound and a level of 80 dB represents the level of pain for humans.

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

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

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

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

Case B : Sound wave in air is the longitudinal wave. so B is not true

The intensity level of sound having intensity I is defined in term of $I _0$. The threshold intensity of hearing is $I _{dB} = 10 \log _{10}({\dfrac{I}{I _0}})$.

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

loudness($L$) in terms of  intensity ($I$) is 

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

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 $1 \ W/m^2.$

headache starts at 80 dB sound for a normal person.

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

A human ear can hear a sound of maximum loudness of $120 \ dB$. Sound intensities above nearly $120 \ dB$ produce pain in the human ear.

Human ear can hear sound from 0dB to 140dB. Intensity of ordinary conversation is rated as nearly 60dB.

Loudness depends on the square of the amplitude of the wave. 

Intensity of reference level  $I _o = 10^{-12} Wm^{-2}$

We know that Loudness of Sound is given by equation,
$\beta  = 10\log \left ( \frac{1}{I _{0}} \right ) dB$
where $I _{0} = 10^{-12} Wm^{-2} $ is the reference unit.
Intensity $ I \propto a^{2} $
where $a$ = amplitude.
Solving we get, the amplitude of the instrument playing music to be reduced by a factor of $100$.

Intensity in $dB$:
$I _{dB} = 10 \log _{10}(\dfrac{I}{I _0})=10 \log _{10}(\dfrac{1}{10^{-12}})=10\times 12 \log _{10}10=120$ $dB$

Intensity in decibels is expressed as $I(dB)=10log(\dfrac{I}{I _0})$

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.

Let $I$ be the  intensity due to a person. For 8 persons, the  intensity, $I^{'}=8I$

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 $100$

Loudness of two sounds are given as  $L _2 = 60 \ dB$ and $L _1 = 30 \ dB$

$90=10 log _{10}(\dfrac{I _1}{I _0})$
$\Rightarrow    \dfrac{I _1}{I _0}=10^9$......(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 
$\dfrac{I _1}{I _2}=10^5$

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

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.

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

The intensity of sound in decibel is given by: 

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

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

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

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

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

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