Two whistles A and B produces notes of frequencies 660 Hz and 594 Hz respectively. There is listener at the mid point of the line joining them. Now the whistle B and the listener start moving with speed 30 m/s towards the whistle A. If speed of sound be 330 m/s, how many beats will be heard by the listener :

Question

Two whistles A and B produces notes of frequencies 660 Hz and 594 Hz respectively. There is listener at the mid point of the line joining them. Now the whistle B and the listener start moving with speed 30 m/s towards the whistle A. If speed of sound be 330 m/s, how many beats will be heard by the listener : (A) 4 (B) 5 (C) 6 (D) 7

Tardigrade Admin · Accepted Answer

As B and listener are moving with the same velocity, frequency of the note from B remains unchanged. Frequency of the note from A, will be n=( ( upsilon - upsilon 1/ upsilon - upsilon s) )n=( (330-30/330-0) )660=600Hz Hence, number of beats heard by the listener is 600-594=6

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7

As B and listener are moving with the same velocity, frequency of the note from B remains unchanged. Frequency of the note from A, will be n=(υ−υs​υ−υ1​​)n=(330−0330−30​)660=600Hz Hence, number of beats heard by the listener is 600−594=6

As B and listener are moving with the same velocity, frequency of the note from B remains unchanged. Frequency of the note from A, will be $n = (\frac{υ - υ _{1}}{υ - υ _{s}}) n = (\frac{330 - 30}{330 - 0}) 660 = 600 Hz$ Hence, number of beats heard by the listener is $600 - 594 = 6$