Question

A source of sound of frequency 256 Hz is moving towards a wall with a velocity of 5m/s. Velocity of sound is 330 m/s. The number of beats heard by an observer standing behind the source is nearly –

• A. $\frac{256 \times 330}{325}–256$
• B. $256–\frac{256 \times 330}{335}$
• C. $\frac{256 \times 330}{325}–\frac{256 \times 330}{335}$
• D. $\frac{256 \times 330}{325}–\frac{256 \times 330}{325}$

Answer 1

Answer: (C)

$\frac{256 \times 330}{325}–\frac{256 \times 330}{335}$

Answer 2

n_D = $256 \left( \frac { 330 } { 330 + 5 } \right)$

n_R = $256 \left( \frac { 330 } { 330 - 5 } \right)$

Dn = n_R – n_D = $\frac { 256 \times 330 } { 325 } - \frac { 256 \times 330 } { 335 }$