PHY 182, Spring Semester 2002

Homework #4: Sound Waves (chapter 19)

Exercise 3:

a) λ = v / f = (343 m/s) / 4.5*10⁶ s^-1 = 76.2 μm

b) λ = v / f = (1500 m/s) / 4.5*10⁶ s^-1 = 333 μm

Exercise 9:

Both waves (P ans S) travel the same distance from the source of the earthquake to the detector of the seismograph

d = v_p*t_p = v_s*t_s with t_s – t_p = 180s.

è t_p = (180 s * v_s) / (v_p – v_s) = (180 s * 4500 m*s^-1) / ((8200-4500) m*s^-1) = 219 s

d = v_p*t_p = 8200 m*s^-1 * 219 s = 1800 km

Exercise 26:

l = 5 m

f = 300 Hz

j₁ - j₂ = p

v(air) = 343 m / s

x₁ = 0 m (position of source 1)

x₂ = l = 5 m (position of source 2)

source 1 emitting from left to right: p₁ = p_m cos (kx - wt)

source 2 emitting from right to left: p₂ = p_m cos (k(x-l) + wt - p)

p = p₁ + p₂ = p_m (cos(kx-wt) + cos(k(x-l)+wt-p))

= p_m (sin(-kx+wt+p/2) + sin(-kx+kl-wt-p/2)) using cosa=sin[(p/2)-a]

= 2 p_m (sin[-kx+(kl/2)]*sin[-(kl/2)+wt+p/2]) using sina + sinb = 2 sin[(a+b)/2]*sin[(a-b)/2]

Maxima in space will occur for

sin[-kx+(kl/2)] = ±1

k((l/2)-x) = n p/2 n = ±1, ±3, ±5, ...

x = (l/2) – (np)/(2k) = (l/2) – (nl/4) = (l/2) – (nv/4f) = 2.5 m – (n*343 m*s^-1)/(4*300 Hz)

n= -1: x = 2.79 m

n = 1: x = 2.21 m

n = -3: x = 3.36 m

n = 3: x = 1.64 m

n = -5: x = 3.93 m

n = 5: x = 1.07 m

n = -7: x = 4.50 m

n = 7: x = 0.50m

or: Maxima occur at ±0.29m, ±0.86m, ±1.43m, ±2.00m from the midpoint.

Exercise 37:

f_beat = 3 Hz

f₁ = ?

f₂ = 384 Hz

f_beat = |f₁-f₂|

Since f decreases when mass (wax) is added, f₁ > f₂ and

f₁ = f_beat + f₂ = 384 Hz + 3 Hz = 387 Hz

Problem 4:

I = P_av / A = (Dp_m)² / (2rv) with Dp_m sin(kx-wt)

a) I_water/I_air = 1 = (Dp_m,water/Dp_m,air)² * r_air * v_air / (r_water*v_water)

Dp_m,water/Dp_m,air = [r_water*v_water / (r_air * v_air)]^-1/2 = [1000*1482/(1.21*343)]^-1/2 = 59.8

b) I_water/I_air = r_air * v_air / (r_water*v_water) = 343*1.21/(1000*1482) = 2.80*10^-4

Problem 6:

Minimum at position 1: pathlength difference SB₁D-SAD = DL₁ = (n+1/2) l n=0,1,2,3,...

Maximum at position 2: pathlength difference SB₂D-SAD = DL₂ = n l n=0,1,2,3,...

For a shift of 1.65 cm at B, the total difference in the pathlength difference is

SB₁D-SB₂D = 2*1.65 cm = 3.30 cm = (n+1/2) l - n l = ½ l

l = 2*3.30 cm = 6.60 cm

a) f = v / l = 343 m*s^-1 / 0.066 m = 5.20 kHz

b) Since I ~ p², p_min²/p_max² = (p_A-p_B)²/(p_A+p_B)² = (10mW/cm²)/ (90mW/cm²)

(p_A-p_B)/(p_A+p_B) = 1/3

ð p_A/p_B = ½

Problem 11:

Resonant frequencies for a closed tube: f_n = n*v/(4L) with n = 1, 3, 5, ...

The distance between successive antinodes is d = l/2, thus f = v*l = 2*d*v

q.e.d.

Problem 15:

Speed of source > speed of light in water ==> Effects at High Speed:

v_s = v/sinq = 0.75 * 3*10⁸ m*s^-1/sin58° = 2.65*10⁸ m/s (>3/4 c)