Question

Two loudspeakers placed 6.0 m apart are driven in phase by an audio oscillator whose frequency range is 2193 Hz to 2967 Hz. A point P is located 4.4 m from one loudspeaker and 3.6 m from the other. The speed of sound is 344 m/s. The frequency produced by the oscillator, for which constructive interference of sound occurs at point P, is closest to

Answer 1

Answer:

"2580 Hz" is the correct solution.

Explanation:

According to the question,

The path difference,

= $4.4 - 3.6$

= $0.80 \ m$

Speed,

= 344 m/s

For constructive interference,

⇒  $Path \ difference =n\times \frac{Speed}{frequency}$

On substituting the values, we get

⇒  $0.80=n\times \frac{344}{frequency}$

⇒  $frequency=n\times 430$

⇒                $n=\frac{frequency}{430}$

If the frequency range is,

f = 2193,

⇒  $n=\frac{2193}{430}$

        $=5.1$

If the frequency range is,

f = 2967,

⇒  $n=\frac{2967}{430}$

        $=6.9$

hence,

For n = 6, the frequency will be:

⇒  $Frequency=n\times 430$

                       $=6\times 430$

                       $=2580 \ Hz$