KEAM 2019 Physics and Chemistry Paper

Case I When train is moving towards a stationary observer.
According to Doppler's effect of sound,
apparent frequency heard, n′=‌

nv

v−vs

where, n= frequency of source, v= speed of sound and vs= speed of source.
⇒‌‌vs=v[1−‌

n

n′

].....(i)
Case II When train is leaving the standing observer.
Again, according to the Doppler's effect of sound, Apparent frequency heard,
n′=‌

nv

v+vs

⇒vs=v[‌

n

n′

−1] .......(ii)
From Eqs. (i) and (ii), we get
‌

n

n′

+‌

n

n′

=2
Given, n′=216Hz and n′=184Hz
Substituting these values in the above equation, we get
n=‌

2×216×184

(216+184)

≈200Hz