Question

Two identical speakers, with the same phase constant, are arranged along a 1D track. One speaker remains at the origin. The other speaker can slide along the track to any position x. You are on the track at x=10 m. You hear interference maxima when the adjustable speaker's position is 0.6 m and 1.2 m and at no points in between. What is the frequency of the sound from the speakers in Hz?

Answer 1

Answer:

$f=571.67\ Hz$

Explanation:

Given:

• Two identical speakers have the same phase constant.

• One of the speaker remains origin.

• The other speaker moves linearly from origin to  x=10m

• position of maxima of the interference, $x_1=0.6\ m$

• position of next maxima of the interference, $x_2=1.2\ m$

Therefore we have the wavelength of the wave as the distance between the two consecutive maxima:

$\lambda=1.2-0.6$

$\lambda=0.6\ m$

We have the speed of sound in the air as:

$v=343\ m.s^{-1}$

Therefore the frequency of sound:

$f=\frac{v}{\lambda}$

$f=\frac{343}{0.6}$

$f=571.67\ Hz$

Answer 2

The frequency of the sound from the speaker is 571.67 Hz.

How do you calculate the frequency?

Given that the phase constant of speakers is the same, the position of one speaker is at origin, the position of another speaker from the origin is 10 m.

The position of maxima interference is at 0.6 m and 1.2 m. Hence the wavelength of the wave is equivalent to the difference between the two consecutive maxima interference.

$\lambda = 1.2 - 0.6$

$\lambda = 0.6 \;\rm m$

We know that the speed v of the sound in the air is 343 m/s. Hence the frequency of the sound is calculated as given below.

$f = \dfrac {v}{\lambda}$

$f = \dfrac { 343}{0.6}$

$f = 571.67\;\rm Hz$

Hence we can conclude that the frequency of the sound from the speaker is 571.67 Hz.

To know more about the frequency, follow the link given below.

https://brainly.com/question/4393505.