A source of the sound of frequency $90$ vibrations/sec is approaching a stationary observer with a speed equal to $1 / 10$ the speed of sound. What will be the frequency heard by the observer

80 v i b r a t i o n s / s e c

No worries! We‘ve got your back. Try BYJU‘S free classes today!

90 v i b r a t i o n s / s e c

No worries! We‘ve got your back. Try BYJU‘S free classes today!

100 v i b r a t i o n s / s e c

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

120 v i b r a t i o n s / s e c

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is D $100 v i b r a t i o n s / s e c$
$n^{'} = n (\frac{v}{v - v_{s}}) = 90 ⎛ ⎜ ⎜ ⎝ \frac{v}{v - \frac{v}{10}} ⎞ ⎟ ⎟ ⎠$
$= 90 \times \frac{10 V}{9 v}$

$= 100 \frac{V i b r a t i o n}{s e c}$

Suggest Corrections

Similar questions

A source of sound of frequency 450 cycles/sec is moving towards a stationary observer with 34 m/sec speed. If the speed of sound is 340 m/sec, then the apparent frequency will be

An observer moves towards a stationary source of sound of frequency n. The apparent frequency heard by him is 2n. If the velocity of sound in air is 332 m/sec, then the velocity of the observer is

Q. A source of sound is moving away from a stationary observer with a speed equal to the speed of sound. The apparent frequency heard by the observer will be:

Q. If to a stationary observer the apparent frequency appear to be 6/5 times the actual frequency of the moving source , then the speed and direction of the sound source is : (velocity of sound = 360 m/sec)

Q. A car has source of sound of frequency

256 Hz

is moving rapidly towards wall with a velocity of

5 m/sec

. How many beats per second will be heard by the driver, if sound travels at a speed of 330 m/sec?

Join BYJU'S Learning Program

A source of the sound of frequency 90 vibrations/sec is approaching a stationary observer with a speed equal to 1/10 the speed of sound. What will be the frequency heard by the observer

The correct option is D 100vibrations/secn′=n(vv−vs)=90⎛⎜ ⎜⎝vv−v10⎞⎟ ⎟⎠=90×10V9v=100Vibrationsec

A source of the sound of frequency $90$ vibrations/sec is approaching a stationary observer with a speed equal to $1 / 10$ the speed of sound. What will be the frequency heard by the observer

The correct option is D $100 v i b r a t i o n s / s e c$
$n^{'} = n (\frac{v}{v - v_{s}}) = 90 ⎛ ⎜ ⎜ ⎝ \frac{v}{v - \frac{v}{10}} ⎞ ⎟ ⎟ ⎠$
$= 90 \times \frac{10 V}{9 v}$

$= 100 \frac{V i b r a t i o n}{s e c}$