A) \[1.5m{{s}^{-1}}\]
B) \[12m{{s}^{-1}}\]
C) \[6\text{ }m{{s}^{-1}}\]
D) \[3\text{ }m{{s}^{-1}}\]
Correct Answer: D
Solution :
When source approaches the observer, the apparent frequency heard by observer is \[n=n\left( \frac{\upsilon }{\upsilon -{{\upsilon }_{s}}} \right)\] ?(1) \[{{\upsilon }_{s}}=\]speed of source of sound During its recession, apparent frequency \[n=n\left( \frac{\upsilon }{\upsilon +{{\upsilon }_{s}}} \right)\] ...(2) Accordingly \[n-n=\frac{2}{100}n\] (given) \[\therefore \] \[n\left( \frac{\upsilon }{\upsilon -{{\upsilon }_{s}}} \right)-n\left( \frac{\upsilon }{\upsilon +{{\upsilon }_{s}}} \right)=\frac{2}{100}n\] or \[\upsilon \left[ \frac{\upsilon +{{\upsilon }_{s}}-\upsilon +{{\upsilon }_{s}}}{(\upsilon -{{\upsilon }_{s}})(\upsilon +{{\upsilon }_{s}})} \right]=\frac{2}{100}\] or \[\frac{2\upsilon {{\upsilon }_{s}}}{(\upsilon -{{\upsilon }_{s}})(\upsilon +{{\upsilon }_{s}})}=\frac{2}{100}\] But speed of sound in air\[\upsilon =300m/s\] \[\therefore \] \[30000{{\upsilon }_{s}}={{(300)}^{2}}-\upsilon _{s}^{2}\] \[\Rightarrow \] \[\upsilon _{s}^{2}+30000\,{{\upsilon }_{s}}-90000\] \[\therefore \]\[\upsilon _{s}^{{}}=\frac{-30000\pm \sqrt{{{(30000)}^{2}}+4\times 9000}}{2}\] \[=-\frac{30000\pm 30006}{2}=\frac{6}{2}=3\,m{{s}^{-1}}\] (taking + ve sign only)
You need to login to perform this action.
You will be redirected in
3 sec
