A) \[1.5\text{ }m{{s}^{-1}}\]
B) \[12\text{ }m{{s}^{-1}}\]
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 \[n=n\left( \frac{\upsilon }{\upsilon -{{\upsilon }_{s}}} \right)\] ... (i) \[{{\upsilon }_{s}}=\] speed of source of sound During its recession, apparent frequency \[n=n\left( \frac{\upsilon }{\upsilon +{{\upsilon }_{s}}} \right)\] ? (ii) 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 \[\frac{2\upsilon \,{{\upsilon }_{s}}}{(\upsilon -{{\upsilon }_{s}})(\upsilon +{{\upsilon }_{s}})}=\frac{2}{100}\] or \[100\,\upsilon {{\upsilon }_{s}}={{\upsilon }^{2}}-\upsilon _{s}^{2}\] But speed of sound in air \[v=300\text{ }m/s\] \[\therefore \] \[30000\,\,{{\upsilon }_{s}}={{(300)}^{2}}-\upsilon _{s}^{2}\] \[\Rightarrow \] \[\upsilon _{s}^{2}+30000\,\,{{\upsilon }_{s}}-90000\] \[\therefore \] \[{{\upsilon }_{s}}=\frac{-300000\pm \sqrt{{{(30000)}^{2}}+4\times 90000}}{2}\] \[=-\frac{300000\pm 30006}{2}=\frac{6}{2}=3\,\,m{{s}^{-1}}\](taking + ve sign only)
You need to login to perform this action.
You will be redirected in
3 sec
