A listener is standing in front of two speakers that are producing sound of the

Question

A listener is standing in front of two speakers that are producing sound of the same frequency and amplitude, except that they are vibrating out of phase. Initially, the distance between the listener and each speaker is the same (see the drawing). As the listener moves sideways, the sound intensity gradually increases. When the distance x in the drawing is 0.97 m, the change reaches the maximum amount (from soft to loud). Using the data shown in the drawing and 343 m/s for the speed of sound, determine the frequency of the sound coming from the speakers 4.36m Out-of-phase speakers

Explanation / Answer

initially the path difference of the sounds from two speakers is 0 for the listener placed symmetrically wrt the speakers
after the motion causing the movement x
path difference, p = sqroot((3.34/2 + x)^2 + 4.36^2) - sqroot((3.34/2)-x)^2 + 4.36^2)
for x = 0.97 m there is maxima at that point
the path difference here should be half of wavelength of the sound as the speakers arte already out of phase
so, using speed of soiund in air, v = 343 m/s
we get lambda = v/f

v/2f = sqroot((3.34/2 + x)^2 + 4.36^2) - sqroot((3.34/2)-x)^2 + 4.36^2)
using values for v and x
343/2f = 0.681144
f = 251.782 z

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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