Question

A stationary source emitting sound of frequency 680Hz is at the origin. An observer is moving with the velocity $\sqrt{2}(\overline{i}+\overline{j})$ m/s at a certain instant. If the speed of sound in air is 340 m/s, then the apparent frequency received by him at that instant is

• A. 680Hz
• B. 676Hz
• C. 684Hz
• D. either 676Hz or 684Hz

Answer 1

The correct option is D either 676Hz or 684Hz

There are two cases :

$v_0=\sqrt{{\left(\sqrt{2}\right)}^2+{\left(\sqrt{2}\right)}^2}=2m/s$

Case I the observer is on $I^{st}$ quadrant

$f^{{}^{\prime }}=680\left(\frac{340-2}{340-0}\right)=676Hz$

Case II the observer is on $I{I}^{nd}$ quadrant

$f^{{}^{\prime }}=680\left(\frac{340+2}{340-0}\right)=684Hz$