Two sound waves with wavelengths 5.0 m and 5.5. m respectively, each propagate in a gas with velocity 330 m/s. We expect the following number of beats per second :-

Question

Two sound waves with wavelengths 5.0 m and 5.5. m respectively, each propagate in a gas with velocity 330 m/s. We expect the following number of beats per second :- (A) 6 (B) 12 (C) 0 (D) 1

Tardigrade Admin · Accepted Answer

Frequency =( text velocity / text wavelength ) ∴ v1=(v/λ1)=(330/5)=66 Hz and v2=(v/λ2)=(330/5.5)=60 Hz Number of beats per second =v1-v2 =66-60=6.

Q. Two sound waves with wavelengths $5.0 m$ and $5.5. m$ respectively, each propagate in a gas with velocity $330 m / s$ . We expect the following number of beats per second :-

6

12

0

1

Frequency $= \frac{velocity}{wavelength}$
$∴ v_{1} = \frac{v}{λ _{1}} = \frac{330}{5} = 66 Hz$
and $v_{2} = \frac{v}{λ _{2}} = \frac{330}{5.5} = 60 Hz$
Number of beats per second $= v_{1} - v_{2}$
$= 66 - 60 = 6$ .

Q. Two sound waves with wavelengths 5.0m and 5.5.m respectively, each propagate in a gas with velocity 330m/s. We expect the following number of beats per second :-

6

12

0

1

Frequency = wavelength velocity ​ ∴v1​=λ1​v​=5330​=66Hz and v2​=λ2​v​=5.5330​=60Hz Number of beats per second =v1​−v2​ =66−60=6.

Q. Two sound waves with wavelengths $5.0 m$ and $5.5. m$ respectively, each propagate in a gas with velocity $330 m / s$ . We expect the following number of beats per second :-

Frequency $= \frac{velocity}{wavelength}$
$∴ v_{1} = \frac{v}{λ _{1}} = \frac{330}{5} = 66 Hz$
and $v_{2} = \frac{v}{λ _{2}} = \frac{330}{5.5} = 60 Hz$
Number of beats per second $= v_{1} - v_{2}$
$= 66 - 60 = 6$ .