Question

A tuning fork vibrates at a frequency of 90 hz and towards an observer with a speed of 0.1 times the speed of sound.what is frequency heard by observer.....solution

Answer 1

Given conditions ⇒

Frequency(f₁) = 90 Hz.

Let the speed of the sound be x m/s.

∴ Speed of the vibrations(v₁) = 0.1x

In Other case,

v₂ = x

Using the Relation,

  v₁/f₁ = v₂/f₂

0.1x/90 = x/f₂

f₂ = 90/0.1

• f₂ = 900 Hz.    

Hope it helps.

Answer 2

this question is based on Doppler Effect, According to Doppler, Apparent frequency can be calculated by formula :

(Apparent Frequency)  $fa = \frac{v+uo}{v-us} fo$

where fa= Apparent Frequency

v = speed of sound

uo = speed of observer (+ve when move towards the source, and -ve when moves away from the source)

us = speed of the source with respect to medium, (+ve when move towards the observer, and -ve when moves away from the observer)

fo = sound frequency

, as per the question,

let v = x then us = 0.1x,

observer is not moving so uo = 0,  and fo = 90Hz

$fa = \frac{x+0}{x-0.1x} 90Hz\\\\= \frac{x}{0.9x} 90Hz\\=100Hz$