Two loudspeakers are placed 1.8 m apart. They play tones of equal frequency. If

Question

Two loudspeakers are placed 1.8 m apart. They play tones of equal frequency. If you stand 3.0 m in front of the speakers, and exactly between them, you hear a maximum of intensity. As you walk parallel to the plane of the speakers, staying 3.0 m away, the sound intensity decreases until reaching a minimum when you arc directly in front of one of the speakers. The speed of sound in the room is 340 m/s. What is the frequency of the sound? Draw, as accurately as you can, a wave-front diagram. On your diagram, label the positions of the two speakers, the point at which the intensity is maximum, and the point at which the intensity is minimum. Use your wave-front diagram to explain why the intensity is a minimum at a point 3.0 m directly in front of one of the speakers.

Explanation / Answer

v = 340 m/s. PUT IN GEOMETRY v = f

but we don’t know f or AC - BC = 0 (0 phase differenc)

AD - BD = /2 ( phase shift)

AD = (3.02+1.82 )^1/2

BD = 3.0

= 2(AD-BD) =1.0 m

Ans- Frequency of sound is 1m

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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