Name: Speed of Sound Wave in Gases
Uploaded: 2019-07-19
Duration: 8 min 7 s
Description: This animation enables students to study Newton’s formula for the sound wave in a gas. Also, the Laplace correction to Newton's formula is more accurate by s...

Question

Two atmospheric sound $A$ and $B$ emit isotropically at constant power. The sound levels $β$ of their emissions are plotted in versus the radial distance $r$ from the sources. The vertical axis scales are set by $β_{1} = 85 .0 dB$ and $β_{2} = 65 .0 dB$ . What are $(a)$ the ratio of the larger power to the smaller power and $(b)$ the sound level difference at $r = 23 m$ ?

Accepted Answer

In the given question, the sources $A$ and $B$ which emit isotropically at constant power.

Given $β_{1} = 85.0$ $dB$ and $β_{2} = 65 .0 dB$

Now, the relation between $(a)$ and power is given by:

$β = 10 \log \frac{P}{I_{0} 4 π r^{2}}$

Now, lets see the difference of $β$ for sounds $A$ and $B$ ,

$∴ β_{A} - β_{B} = Δ β = 10 \log \frac{P_{A}}{I_{0} 4 π r^{2}} - 10 \log \frac{P_{B}}{I_{0} 4 π r^{2}}$

$\Rightarrow β = 10 \log \frac{P_{A}}{\frac{I_{0} 4 π r^{2}}{\frac{P_{B}}{I_{0} 4 π r^{2}}}}$ $[\log A - \log B = \frac{\log A}{\log B}]$

$Δ β \Rightarrow 10 \log \frac{P_{A}}{P_{B}}$

Now, from the above relation, we cancelled that the $Δ β$ term is independent of $' r'$ and can be calculated everywhere.

$(a)$ Now, from the given diagram we can se that, $Δ β = 5 dB$ .

$Δ β = 10 \log \frac{P_{A}}{P_{B}}$

$5 = 10 \log \frac{P_{A}}{P_{B}}$

$\frac{5}{10} = \log \frac{P_{A}}{P_{B}}$

$\frac{1}{2} = \log \frac{P_{A}}{P_{B}}$

$0 .5 = \log \frac{P_{A}}{P_{B}}$

$\frac{P_{A}}{P_{B}} = {(10)}^{0 .5} = 3 : 16$

Therefore, in this part, ratio of powers, $P_{A} : P_{B} = 3 : 16$

$(b)$ In this part we have to calculate the sound level at, $r = 23 m$

As discussed earlier, the sound level is independent of $r$ . Also, from the diagram we can see, for every position, the distance between them is constant.

Important Questions on Waves-II

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