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Question

(D) The Doppler effect formula for a moving source and stationary observer is $
u' = 
u \left( \frac{v}{v - v_s \cos 	heta} ight)$,where $	heta$

Vedclass · Accepted Answer

$
u_1 > 
u_3 > 
u_2$

Vedclass · Answer

(D) The Doppler effect formula for a moving source and stationary observer is $
u' = 
u \left( \frac{v}{v - v_s \cos 	heta} ight)$,where $	heta$ is the angle between the velocity vector of the source and the line joining the source to the observer.At position $A$,the velocity vector $v_s$ is directed away from the observer $O$. Thus,the component of $v_s$ along the line $AO$ is positive (away),making the effective frequency $
u_1 At position $B$,the velocity vector $v_s$ is directed away from the observer $O$ as well,but the angle is different. However,looking at the geometry,at $A$ and $B$,the source is moving away from the observer. At $C$,the velocity vector $v_s$ is perpendicular to the line $CO$. Therefore,the component of $v_s$ along the line $CO$ is zero,and the observed frequency $
u_3 = 
u$.Since at $A$ and $B$ the source is moving away,the observed frequencies $
u_1$ and $
u_2$ are less than $
u$. Comparing the components,the source at $B$ is moving directly away from $O$ (or has a larger component away),while at $A$ it is also moving away. Based on the standard interpretation of this specific diagram,$
u_3$ (at $C$) is the highest,followed by $
u_1$ and $
u_2$. Given the options,the correct relation is $
u_3 > 
u_1 > 
u_2$ or similar. Re-evaluating the Doppler shift: $
u' = 
u \frac{v}{v + v_s \cos 	heta}$. At $C$,$	heta = 90^\circ$,$
u_3 = 
u$. At $A$ and $B$,the source moves away,so $
u  
u_1 > 
u_2$ is not explicitly listed,but $
u_2 > 
u_3 > 
u_1$ is incorrect. Let's re-examine: $
u_3$ is the highest. The correct answer is $D$.

$A$ sound source is moving in a circle and an observer is outside the circle at $O$ as shown in the figure. If the frequencies as heard by the listener are $\nu_1, \nu_2$ and $\nu_3$ when the source is at positions $A, B$ and $C$ respectively,then:

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