PHY 182, Spring Semester 2002

Homework #2: Waves (chapter 18)

Exercise 7:

a) y_m = 6.0 cm = 0.06 m

b) λ = 2π / k = 2π / (2π m^-1) = 1.0 m

c) f = ω / (2π) = 4π / (2π s^-1) = 2.0 Hz

d) v_x = λ*f = 1.0 m * 2.0 s^-1 = 2.0 m /s

e) negative direction, as waves of the form y(x,t) = f(x+vt) travel in the negative direction, y(x,t) = f(x-vt) in the positive direction

f) u_y = dy / dt = - ω*y_m*cos(kx-ωt) is maximal for cos(kx-ωt) = ± 1

u_y,max = ω*y_m = 4π s^-1 * 0.06 m = 0.75 m / s

Exercise 11:

a) From the picture: y_m = 0.05 m

b) From the picture: λ = 0.4 m

c) v_x = (F / μ)^-1/2 = (3.6 N / (0.025 kg m^-1)^-1/2 = 12 m / s

d) T = λ / v_x = 0.4 m / (12 m/s) = 0.033 s

e) u_y,max = ω*y_m = (2π/T)*y_m = (2π/0.033s)*0.05m = 3π m / s = 9.4 m /s

f) y(x,t) = y_m sin(kx-ωt-φ)

y(0,0) = 0.04m = y_m(4/5) è sin(-φ) = 4/5 è φ = -arcsin(4/5) = -0.30π

k = 2π / λ = (2π / 0.4) m^-1 = 16 m^-1

ω = 2π / T = (2π / 0.033) s^-1 = 190 s^-1

è y(x,t) = (0.05 m) sin(16 m^-1*x + 190 s^-1*t + 0.30π)

Exercise 27 :

a) y(2.3m,0.16s) = (0.15m) sin[(0.79rad/s)*2.3m – (13 rad/s)*0.16s] =

-0.039m

b) y(x,t) = (0.15m) sin[(0.79rad/s)*x + (13 rad/s)*t]

c) y_s(x,t) = (0.15m) sin[(0.79rad/s)*x + (13 rad/s)*t] + (0.15m) sin[(0.79rad/s)*x + (13 rad/s)*t]

y_s(2.3m,0.16s) = (0.15m) sin[(0.79rad/s)*2.3m - (13 rad/s)*0.16s] + (0.15m) sin[(0.79rad/s)*2.3m + (13 rad/s)*0.16s] = -0.14 m

Problem 4:

y(0.096m,t) = (0.0512m) sin[1.16 – (4.08 s^-1) t]

v_x = 0.826 m /s

m = 0.386 kg / m

a) f = w / 2p = 4.08 s^-1 / 2p = 0.65 Hz

b) l = v_x T = v_x / f = (0.826 m /s) / 0.65 s^-1 = 1.27 m

c) k = 2p / l = 2p / (1.27 m) = 4.95 m^-1

w = 2p f = 2p 0.65 Hz = 4.08 s^-1

y(0.096m,0) = (0.0512m) sin[1.16]

è 1.16 = kx - f

f = kx –1.16 = 4.95 m^-1 * 0.096 m – 1.16 = -0.685

y(x,t) = (0.0512m) sin[4.95 m^-1 x - 4.08 s^-1 t + 0.685]

d) F = m v_x² = (0.386 kg / m) (0.826 m / s)² = 0.264 N

Problem 14:

1: in-phase:

2[H² + (d/2)²]^1/2 – d = n l with n = 0, 1, 2, 3, ...

2: out-of-phase:

2[(H+h)² + (d/2)²]^1/2 – d = (n+1/2) l = n l + ½ l with n = 0, 1, 2, 3, ...

Subtract 1 from 2: l = 4[(H+h)² + (d/2)²]^1/2 – 4[H² + (d/2)²]^1/2

Problem 21:

m = 0.0442 kg

L = 0.924 m

f_n = 60.0 Hz

n = 4

f_n = (n/2) L v = (n/2L) (F/ m)^-1/2 = (n/2L) (F*L / m)^-1/2

è F = (4 f_n² m L) / n² = (4*60.0² s^-2 0.0442 kg 0.924 m) / 4² = 36.8 N