Looking for classes? Ksquare Career Institute, Bengaluru →
Two stationary sources each emitting waves of wavelength $\lambda$. An observer moves from one source to other with velocity u. Then number of beats heard by him :
A
$\dfrac{2u}{\lambda}$
B
$\dfrac{u}{\lambda}$
C
$u\lambda$
D
$\dfrac{u}{2\lambda}$
Detailed Solution
[add image: observer moving with speed u from source $S_1$ towards source $S_2$]
The observer moves away from the first source: $n_1 = n\left(\dfrac{v - u}{v}\right) = \left(1 - \dfrac{u}{v}\right)n$
and towards the second source: $n_2 = n\left(\dfrac{v + u}{v}\right) = \left(1 + \dfrac{u}{v}\right)n$
Beat frequency $= |n_2 - n_1| = \dfrac{2nu}{v}$
Since $v = n\lambda$, $\dfrac{n}{v} = \dfrac{1}{\lambda}$: beats $= \dfrac{2u}{\lambda}$
The observer moves away from the first source: $n_1 = n\left(\dfrac{v - u}{v}\right) = \left(1 - \dfrac{u}{v}\right)n$
and towards the second source: $n_2 = n\left(\dfrac{v + u}{v}\right) = \left(1 + \dfrac{u}{v}\right)n$
Beat frequency $= |n_2 - n_1| = \dfrac{2nu}{v}$
Since $v = n\lambda$, $\dfrac{n}{v} = \dfrac{1}{\lambda}$: beats $= \dfrac{2u}{\lambda}$