Ch 16 #pulse, 15, 28, 40, 46, bonus

Ch 16.1 pulse

A t = 0, a transverse pulse in a string is described by the function, y = x/(x² + 5), using SI units. Write the function y(x, t) that describes this pulse if it is traveling in the positive x-dir with a speed of 6 m/s.

If traveling in the positive direction we know the equation in the x-dir is in the form of: x – vt; where v = 6 So x – 6t describes our x-dir motion.

y(x,t) = x / [ x^2 + 5] y(x,t) = x / [(x-6t)^2 + 5]

Ch 16.2 #15

(a) Write the expression for y as a function of x and t, for a sinusoidal wave traveling along a rope in the –x with the following characteristics: A = 8.00 cm, λ = 80.0 cm, f = 3.00 Hz, and y(0, t) = 0 at t = 0. (b) Write an expression for y as a function of x and t for the wave in part (a) assuming that y(x, 0) = 0 at the point x = 10.0 cm.

(a) k = 2π/λ k = 2π/0.800 k = 7.85 m^-1	ω = 2πf ω = 2π3 ω = 6π rad/sec	(b) We know that we must shift the wave so that y(0.100, 0) = 0 y(x,t) = 0.0800 sin(7.85x + 6πt + φ) 0 = 0.08 sin (7.85(0.1) + 0 + φ) 0 = sin (.785 + φ) φ = -0.785
y(x,t) = 0.0800 sin(7.85x + 6πt) meters Remember y(0, 0) must equal 0

Ch 16.3 #28

A simple pendulum consists of a ball of mass M hanging from a uniform string of mass m and length L, with m << M. If the period of oscillations for the pendulum is T, determine the speed of a transverse wave in the string when the pendulum hangs at rest.

v^2 = FT / μ v^2 = (Mg/m) L v^2 = (Mg/m) T^2g / 4π^2 v = ½gT√M / π√m

μ = m/L T = 2π(L/g)^1/2 L = T^2 g / 4π^2

Ch 16.5 #40

The wave function for a wave on a taut string is y(x, t) = (0.350 m) sin(10πt – 3πx + π/4) where x is in meters and t in seconds. (a) What is the average rate at which energy is transmitted along the string if the linear mass density is 75.0 g/m? (b) What is the energy contained in each cycle of the wave? Ans. 15.1 W, (b)3.02 J

y(x, t) = (0.350 m) sin(10πt – 3πx + π/4) y(x, t) = A sin(ω t – k x + φ )			(a) P = ½ μ ω^2 A^2 v P = ½ 0.075 100π^2 0.35^2 3.33 P = 15.1 Watt
A = 0.350 m ω = 10π rad/s k = +3π m^-1 φ = π/4	ω = 2πf f = 10π/2π f = 5 Hz	k = 2π/λ λ = 2π/k λ = 2π/3π λ = 2/3 m	(b) P = Work / time P = E / (λ /v); v = λ / T E = P (λ /v) E = ½ μ ω^2 A^2 v (λ /v) E = ½ μ ω^2 A^2 λ E = ½ 0.075 100π^2 0.35^2 (2/3) E = 3.02 Joules
v = f λ v = 5(2/3) v = 3.33 m/s

Ch 16.6 #46

(a) Show that the function y(x,t) = x² + v²t² is a solution to the wave equation. (b) Show that the function in part (a) can be written as f(x + vt) + g(x - vt), and determine the functional forms for f and g. (c) Repeat parts a and b for the function y(x, t) = sin(x) cos(vt)

Note to grader: Don’t grade Part A…this was given.

y(x,t) = x^2 + v^2t^2 ∂y/∂x = 2x ∂^2y/∂x^2 = 2	y(x,t) = x^2 + v^2t^2 ∂y/∂t = 2v^2t ∂^2y/∂t^2 = 2v^2	(b) f(x + vt) + g(x - vt) ½(x + vt)^2 + ½(x - vt)^2 ½x^2 + ½2xvt+ ½v^2t^2 + ½x^2 - ½2xvt + ½v^2t^2 f(x + vt) + g(x - vt) = x^2 + v^2t^2 f(x + vt) = ½(x + vt)^2 g(x - vt) = ½(x - vt)^2
(the below is cut and pasted from our online notes) The linear wave equation for a string is (μ/T) ∂^2y/∂t^2 = ∂^2y/∂x^2 which is also ∂^2y/∂x^2 = (1/v^2) ∂^2y/∂t^2 ∂^2y/∂x^2 = 2 and ∂^2y/∂x^2 = 2v^2 2 = (1/v^2) 2v^2 2 = 2 Thus y(x,t) = x^2 + v^2t^2 is a solution to the wave equation.
(c) y(x,t) = sin(x) cos(vt) ∂y/∂x = cos(x) cos(vt) ∂^2y/∂x^2 = -sin(x) cos(vt)	y(x,t) = sin(x) cos(vt) ∂y/∂t = -vsin(x) sin(vt) ∂^2y/∂t^2 = -v^2sin(x) cos(vt)	(c) use trig identity sin(a+b) = sin(a)cos(b) + cos(a)sin(b) sin(x+vt) = sin(x)cos(vt) + cos(x)sin(vt) sin(x-vt) = sin(x)cos(vt) - cos(x)sin(vt) sin(x+vt) + sin(x-vt) = 2sin(x)cos(vt) f(x + vt) + g(x - vt) = sin(x)cos(vt) ½sin(x+vt) + ½ sin(x-vt) = sin(x)cos(vt)
∂^2y/∂x^2 = (1/v^2) ∂^2y/∂t^2 -sin(x) cos(vt) = (1/v^2) -v^2sin(x) cos(vt) -sin(x) cos(vt) = -sin(x) cos(vt) Thus y(x,t) = sin(x) cos(vt) is a solution to the wave equation.

Bonus Wave speed

Calculate the speed of longitudinal waves along a spring of force constant, k, and unstretched length of the spring is L, and μ is the mass per unit length.

Ans: v² = kL/μ