..ooo T-Mobile 10:01 PM gannon.blackboard.com C Sound Name: Name: Name: Two spea

Question

..ooo T-Mobile 10:01 PM gannon.blackboard.com C Sound Name: Name: Name: Two speakers are separated by a distance 1.5 m and emit sound waves in phase. A listener is seated -3.5m away directly in front of one speaker. The speed of sound in air is v=343 m/s 1) What is the lowest frequency that will cause constructive interference? 2) What is the lowest frequency that will cause destructive interference? 3) The second speaker is disconnected (so sound only comes from one speaker) Ata distance of 1.0m, the sound level is 85 dB. What is the sound intensity 15 m from the speaker?

Explanation / Answer

a)

The path difference between the waves = sqrt(L^2 + d^2) - L

= sqrt(3.5^2 + 1.5^2) - 3.5

= 0.30789 m

Now, Let wavelength of the waves be K

So, for constructive interference, path difference = nK

where n = any interger starting from 0

So, for lowest frequency, n = 1

So, 0.30789 = K

So, frequency , f = v/K = 343/0.30789 = 1114 Hz

b)

The path difference between the waves = sqrt(L^2 + d^2) - L

= sqrt(3.5^2 + 1.5^2) - 3.5

= 0.30789 m

Now, Let wavelength of the waves be K

So, for destructive interference, path difference = nK + K/2

where n = any interger starting from 0

So, for lowest frequency, n = 0

and K/2 = 0.30789

So, K = 0.61578 m

So, frequency = v/K = 343/0.61578 = 557 Hz

c)

I2 / I1 = (r1 / r2)^2

So, 10*log( I2 / I1 ) = 20*log(r1/ r2)

Now, 10*log(I2) - 10*log(I1) = 20*log(r1 / r2)

So, 10*log(I2) - 85 = 20*log(1/15)

So, 10*log(I2) = 61.5 dB <--------- answer

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.