Browse by Stream
Discuss Doubts
Home
Search for Exam, Articles, Questions
get app
Go To Your Study Dashboard
  1. Home
  2. Solve this problem The wavelength is 600 cm when the source is stationary. If the source is moving with relative velocity of 60 m/sec towards the observer, then the wavelength of the sound wave reaching to the observer will be (velocity of sound = 33
# Engineering

The wavelength is 600 cm when the source is stationary. If the source is moving with relative velocity of 60 m/sec towards the observer, then the wavelength of the sound wave reaching to the observer will be (velocity of sound = 330 m/s)

  • Option 1)

    98 cm

  • Option 2)

    60 cm

  • Option 3)

    49 cm

  • Option 4)

    120 cm

 

Answers (1)
V Vakul

As we learnt in 

Frequency of sound when observer is stationary and source is moving towards observer -

\nu {}'= \nu _{0}.\frac{C}{C-V_{s}}
 

- wherein

C= speed of sound

V_{s}= speed of source

\nu _{0}= original frequency

\nu {}'= apparent frequency

 

 \nu=\nu_{o}\:.\:(\frac{C}{C-V_{s}})\:\:\:\therefore\frac{\nu}{\nu_{o}}=\frac{C}{C-V_{s}}

\lambda\:\propto \frac{1}{\nu}\:\:\:=>\frac{\lambda_{2}}{\lambda_{1}}=\frac{C-V_{s}}{C}=\frac{330-60}{330}=\frac{270}{330}

\lambda_{2}=\frac{9}{11}\times 60.0\:cm\:=\:49\:cm

 


Option 1)

98 cm

This option is incorrect.

Option 2)

60 cm

This option is incorrect.

Option 3)

49 cm

This option is correct.

Option 4)

120 cm

This option is incorrect.

Similar Questions

Preparation Products

Knockout BITSAT 2021

It is an exhaustive preparation module made exclusively for cracking BITSAT..

₹ 4999/- ₹ 2999/-
Buy Now
Knockout BITSAT-JEE Main 2021

An exhaustive E-learning program for the complete preparation of JEE Main and Bitsat.

₹ 27999/- ₹ 16999/-
Buy Now
Exams
Articles
Questions