P18_029.PDF

29. (a) When the right side of the instrument is pulled out a distance d the path length for sound waves

increases by 2 d . Since the interference pattern changes from a minimum to the next maximum, this

distance must be half a wavelength of the sound. So 2 d = λ/ 2, where λ is the wavelength. Thus

λ =4 d and, if v is the speed of sound, the frequency is f = v/λ = v/ 4 d = (343m / s) / 4(0 . 0165m) =

5 . 2

Write √ I = Cs m ,where I is the intensity, s m is the displacement amplitude, and C is a constant

of proportionality. At the minimum, interference is destructive and the displacement amplitude is

the diﬀerence in the amplitudes of the individual waves: s m = s SAD − s SBD , where the subscripts

indicate the paths of the waves. At the maximum, the waves interfere constructively and the

displace men t amplitude is the sum of the am plitudes of the individual waves: s m = s SAD + s SBD .

Solve √ 100 = C ( s SAD −

s SBD )an d √ 900 = C ( s SAD + s SBD )for s SAD and s SBD .Addthe

eq uati ons t o o btain s SAD =( √ 100 + √ 900) / 2 C =20 /C , then subtract them to obtain s SBD =

− √ 100) / 2 C =10 /C . The ratio of the amplitudes is s SAD /s SBD =2.

result in a decrease in the displacement amplitude. Those losses are greater on path B since it is

longer than path A.

10 3 Hz.

(b) The dis pl acement amplitude is proportional to the square root of the intensity (see Eq. 18–27).

( √ 900