A) \[\frac{19}{18}\]
B) \[\frac{18}{19}\]
C) \[\frac{1}{2}\]
D) \[2\]
Correct Answer: A
Solution :
Key Idea: The perceived frequency of observer will depend on the relative motion of observer and source (train). The perceived frequency by the observer \[f={{f}_{0}}\left( \frac{v\pm {{v}_{0}}}{v\pm {{v}_{s}}} \right)\] Since, observer is at rest and source is moving towards observer, the above relation assumes the form \[f={{f}_{o}}\left( \frac{v}{v-{{v}_{s}}} \right)\] In 1st case: \[v=340\,m/s,{{v}_{s}}=34\,m/s\] \[\therefore \] \[{{f}_{1}}={{f}_{0}}\left( \frac{340}{340-34} \right)\] or \[{{f}_{1}}=\frac{340}{306}{{f}_{o}}\] ? (i) In 2nd case: \[v=340\,\,m/s,\,{{v}_{s}}=17\,m/s\] \[\therefore \] \[{{f}_{2}}={{f}_{0}}\left( \frac{340}{340-17} \right)\] or \[{{f}_{2}}=\frac{340}{323}{{f}_{o}}\] ?(ii) Dividing Eq. (i) by Eq. (ii), we obtain \[\frac{{{f}_{1}}}{{{f}_{2}}}=\frac{323}{306}=\frac{19}{18}\]
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