Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • How to solve this problem- Two passenger trains moving with a speed of 108 km/hour cross each other. One of them blows a whistle whose frequency is 1500 Hz. If sound speed is 330 m/s, then passengers sitting in the other train, after trains cross eac

Two passenger trains moving with a speed of 108 km/hour cross each other. One of them blows a whistle whose frequency is 1500 Hz. If sound speed is 330 m/s, then passengers sitting in the other train, after trains cross each other will hear sound whose frequency will be

  • Option 1)

    1250 Hz

  • Option 2)

    625 Hz

  • Option 3)

    1500 Hz

  • Option 4)

    900 Hz

 

Answers (1)

best_answer

As we learnt in 

Frequency of sound when source & observer are moving away from each other -

\nu {}'= \nu _{0}.\frac{C-V_{0}}{C+V_{s}}
 

- wherein

C= Speed of sound

V_{0}= Speed of observer

V_{s}= speed of source

\nu _{0 }= Original frequency

\nu {}'= apparent frequency

 

 \nu_{o}\:=\:1500

Here both are moving away from each other

\Rightarrow \nu=\nu_{o}\:.\:\left(\frac{C-V_{o}}{C+V_{s}} \right )=1500\:\left(\frac{330-30}{330+30} \right )

\because\:V=108\:Km/hr\:=\:30\:m/s

\nu=1500\times\frac{300}{360}\:Hz\:=1250\:Hz

 


Option 1)

1250 Hz

This option is correct.

Option 2)

625 Hz

This option is incorrect.

Option 3)

1500 Hz

This option is incorrect.

Option 4)

900 Hz

This option is incorrect.

Posted by

Plabita

View full answer

Similar Questions

Trending Articles/News

anam-khan

BITSAT 2025 session 2 registration deadline today; correction facility from June 13
anam-khan

BITS Pilani begins direct admission process 2025 for board exam toppers; eligibility, tie-breaking rule
anam-khan

BITSAT 2025 session 1 absentee slot registration begins; exam on June 2

Latest Question