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The Sound Level at a Point 5.0 M Away from a Point Source is 40 Db. What Will Be the Level at a Point 50 M Away from the Source? - Physics

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Question

The sound level at a point 5.0 m away from a point source is 40 dB. What will be the level at a point 50 m away from the source?

Sum
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Solution

Let

\[\beta_A\]  be the sound level at a point 5 m (= r1) away from the point source and 

\[\beta_B\]be the sound level at a distance of 50 m (= r2) away from the point source.
∴​\[\beta_A\]= 40 dB
Sound level is given by:
\[\beta = 10 \log_{10} \left( \frac{I}{I_0} \right)\]
According to the question,

\[\beta_A  = 10   \log_{10}   \left( \frac{I_A}{I_0} \right) . \] 

\[ \Rightarrow   \frac{I_A}{I_0} =  {10}^\left( \frac{\beta_A}{10} \right)        .  .  .  .  . \left( 1 \right)\] 

\[ \beta_B  = 10   \log_{10} \left( \frac{I_B}{I_o} \right)\] 

\[ \Rightarrow   \frac{I_B}{I_0} =  {10}^\left( \frac{\beta_B}{10} \right)        .  .  .  .  . \left( 2 \right)\] 

\[\text { From } \left( 1 \right) \text { and } \left( 2 \right),   \text { we  get: }\] 

\[          \frac{I_A}{I_B} =  {10}^\left( \frac{\beta_A - \beta_B}{10} \right)        .  .  .  . \left( 3 \right)\] 

\[\text { Also }, \] 

\[  \frac{I_A}{I_B} = \frac{r_B^2}{r_A^2} =  \left( \frac{50}{5} \right)^2    =    {10}^2        .  .  .  .  . \left( 4 \right)\] 

\[\text { From } \left( 3 \right)  \text { and } \left( 4 \right), \text{  we  get: }\] 

\[ {10}^2  =  {10}^\left( \frac{\beta_A - \beta_B}{10} \right) \] 

\[ \Rightarrow   \frac{\beta_A - \beta_B}{10} = 2  \] 

\[ \Rightarrow    \beta_A  -  \beta_B  = 20\] 

\[ \Rightarrow    \beta_B  = 40 - 20 = 20  dB\]

\[\beta_A  = 10   \log_{10}   \left( \frac{I_A}{I_0} \right) . \] 

\[ \Rightarrow   \frac{I_A}{I_0} =  {10}^\left( \frac{\beta_A}{10} \right)        .  .  .  .  . \left( 1 \right)\] 

\[ \beta_B  = 10   \log_{10} \left( \frac{I_B}{I_o} \right)\] 

\[ \Rightarrow   \frac{I_B}{I_0} =  {10}^\left( \frac{\beta_B}{10} \right)        .  .  .  .  . \left( 2 \right)\] 

\[\text { From }\left( 1 \right) \text{ and } \left( 2 \right), \text  { we  get:  }\] 

\[          \frac{I_A}{I_B} =  {10}^\left( \frac{\beta_A - \beta_B}{10} \right)        .  .  .  . \left( 3 \right)\] 

\[\text { Also }, \] 

\[  \frac{I_A}{I_B} = \frac{r_B^2}{r_A^2} =  \left( \frac{50}{5} \right)^2    =    {10}^2        .  .  .  .  . \left( 4 \right)\] 

\[\text { From } \left( 3 \right)  \text { and } \left( 4 \right),   \text { we  get: } \] 

\[ {10}^2  =  {10}^\left( \frac{\beta_A - \beta_B}{10} \right) \] 

\[ \Rightarrow   \frac{\beta_A - \beta_B}{10} = 2  \] 

\[ \Rightarrow    \beta_A  -  \beta_B  = 20\] 

\[ \Rightarrow    \beta_B  = 40 - 20 = 20  dB\]

Thus, the sound level of a point 50 m away from the point source is 20 dB.

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Wave Motion
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Chapter 16: Sound Waves - Exercise [Page 353]

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HC Verma Concepts of Physics Vol. 1 [English] Class 11 and 12
Chapter 16 Sound Waves
Exercise | Q 19 | Page 353

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