Two tuning forks having frequencies 460 Hz and 454 Hz are tuned together. The time interval between successive maximum intensity will be

Question

Two tuning forks having frequencies 460 Hz and 454 Hz are tuned together. The time interval between successive maximum intensity will be (A) (1/6) s (B) (1/2) s (C) 1 s (D) 4 s

Tardigrade Admin · Accepted Answer

Number of beats/second =v1-v2=460-454=6. ∴ Time interval between two successive beats =(1/6) s This is the time interval between successive maximum intensity.

Q. Two tuning forks having frequencies $460 Hz$ and $454 Hz$ are tuned together. The time interval between successive maximum intensity will be

$\frac{1}{6} s$

$\frac{1}{2} s$

1 s

4 s

Number of beats/second $= v_{1} - v_{2} = 460 - 454 = 6$ .
$∴$ Time interval between two successive beats $= \frac{1}{6} s$
This is the time interval between successive maximum intensity.

Q. Two tuning forks having frequencies 460Hz and 454Hz are tuned together. The time interval between successive maximum intensity will be

61​s

21​s

1 s

4 s

Number of beats/second =v1​−v2​=460−454=6. ∴ Time interval between two successive beats =61​s This is the time interval between successive maximum intensity.

Q. Two tuning forks having frequencies $460 Hz$ and $454 Hz$ are tuned together. The time interval between successive maximum intensity will be

$\frac{1}{6} s$

$\frac{1}{2} s$

Number of beats/second $= v_{1} - v_{2} = 460 - 454 = 6$ .
$∴$ Time interval between two successive beats $= \frac{1}{6} s$
This is the time interval between successive maximum intensity.