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Section: Physics

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Question : 11 of 31

Marks: +1, -0

A train S1, moving with a uniform velocity of 108 km/h approaches another train S2 standing on a platform. An observer O moves with a uniform velocity of 36km/h towards S2 as shown in figure. Both the trains are blowing whistles of same frequency 120Hz. When O is 600m away from S2 and distance between S1 and S2 is 800m, the number of beats heard by O is_____

[ Speed of the sound =330m/s]

[JEE Adv 2019 P1]

Your Answer:

Solution:

\tan\theta_{1} = \frac{800}{600} = \frac{4}{3}

\theta_{1}=53^{\circ}

\theta_{2}=37^{\circ}

⇒ Freq. Heared form train at

S_2=f_2

f_{2}=\left(\frac{V+V_0}{V}\right)f_S

=\left(\frac{330+10}{330}\right)\times f_s

=\frac{340}{330}f_s=1.030f_s

⇒freq. Heared from train at

S_1=f_1

f_{1}=\left(\frac{V+V_0\cos\theta_1}{V-V_s\cos\theta_2}\right)f_s

=\left(\frac{330+10\times\frac{3}{5}}{330-30\times\frac{4}{5}}\right)f_s

f_{1}=\left(\frac{336}{306}\right)f_s=1.098f_s

⇒ Beat fieq

=1.098f_s-1.030f_s

=0.068f_s

=0.068\times120

=8.1

6

\cong 8

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