AP EAMCET201822 Apr 2018Evening ShiftPhysicsWaves and SoundActual
An observer and a source emitting sound of frequency 120 ~Hz are on the X -axis. The observer is stationary while the source of sound is in motion given by the equation x=3 t ( x is in metres and t is in seconds). If the difference between the maximum and minimum frequencies of the sound observed by the observers is 22 ~Hz , then the value of is ( ppeed of sound in air =330 ~ms ⁻¹ )
Options
- A33 rad s ⁻¹
- B36 rad s ⁻¹
- C20 rad s ⁻¹
- D10 rad s ⁻¹
Correct answer
D. 10 rad s ⁻¹
Step-by-step solution
Instantaneous speed of source is v= d x d t =3 t Difference between maximum and minimum frequencies is 22 ~Hz . So, f_ -f_ =f ( v v-v_s )-f ( v v+v_s )=22 Now here, f=120 ~Hz , v=330 ~ms ⁻¹, v_s=3 Substituting these values in Eq (i), we get 120 ( 330 330-3 - 330 330+3 )=22 =10 ~s ⁻¹