Question

A train is travelling at 120 kmph and blows a whistle of frequency 1000 Hz. The frequency of the note heard by a stationary observer, if the train is approaching him and moving away from him, are: (Velocity of sound in air = 330ms${}^{-1}$)

• A. 1112Hz, 908Hz
• B. 908Hz, 1112Hz
• C. 1080Hz, 820Hz
• D. 820Hz,1080Hz

Answer 1

Answer: (A)

1112Hz, 908Hz

Given : $v_0=0m/s$

$V=330m/s$

$\delta =1000H_z$

$V_s=120km/h=120\times \frac{1000}{3600}=\frac{100}{3}m/s$ (train approach)

$V_s=-120km/h=\frac{100}{3}m/s$ (train moves away)

By Doppler effect formula

${\delta }_1^1=\delta \left(\frac{v-v_0}{v-v_s}\right)$

$=1000\left(\frac{330-0}{330-\frac{100}{3}}\right)$

$\frac{1000\times 330\times 3}{890}$

$=1112H_z$

${\delta }_2^1=\delta \left(\frac{v-v_0}{v-v_s}\right)$

$=1000\left(\frac{330-0}{330+\frac{100}{3}}\right)$

$=\frac{1000\times 330\times 3}{1090}$

$=908H_z$