Ch 18 #6, 15, shoe, sis, 45, 49, guitar

Ch 18.1 #6

Two identical sinusoidal waves with wavelengths of 3.00 m travel in the same direction at a speed of 2.00 m/s. The second wave originates from the same point as the first, but at a later time. Determine the minimum possible time interval between the starting moments of the two waves if the amplitude of the resultant wave is the same as that of each of the two initial waves.

y = 2Acos(φ/2) sin(kx-wt+φ/2) Given: 2A cos (φ /2) = A

2 cos (φ /2) = 1 (φ /2) = cos^-1(1/2) Phase difference of φ = 120° or 1/3 of a wave

v = f λ v = 2; λ = 3 f = 2/3 Hz T = 1.5 seconds 1/3 of the period is ½ seconds

Ch 18.2 #15

Two speakers are driven in phase by a common oscillator at 800 Hz and face each other at a distance of 1.25 m. Locate the points along a line joining the two speakers where relative minima of sound pressure amplitude would be expected. (Use v = 343 m/s.)

Two antinodes are required for a full wavelength in standing waves which are produced by the facing speakers: dNtoN = λ/2 or λ =2dNtoN v = f λ v = f 2dNtoN 343 = 800(2dNtoN) dNtoN = 0.214 meters If the speakers vibrate in phase, the point halfway between them is an antinode of pressure at a distance from either speaker of 1.25 m / 2 = 0.625m

Then there is a node at 0.625 – ½0.214 = 0.518 m a node at 0.518 – 0.214 = 0.303 m a node at 0.303 – 0.214 = 0.0891 m a node at 0.518 + 0.214 = 0.732 m a node at 0.732 + 0.214 = 0.947 m & a node @ 0.947 + 0.214 = 1.16 m from either speaker.

Ch 18.3 shoe

A concrete shoe (ρ_concrete = 2.50 g/cc) is hung from scale by a wire, while being completely submerged in water. This wire emits a fundamental frequency of 500 Hz when plucked. The water is replaced with oil (ρ_oil = 0.900 g/cc). What is the new fundamental frequency?

FT-init = ρconcreteVg – ρwaterVg FT-init = mg - FBouy FT-init = 2.50Vg – 1.00Vg FT-init = 1.5Vg FT-final = ρconcreteVg – ρoilVg FT-final = 2.5Vg – 0.9Vg FT-final = 1.6Vg	f^2 λ^2 = FT / μ f^2 = FT / μλ^2	finit^2 = FT-init / μλ^2 ffinal^2 = FT-final / μλ^2
	finit^2 /ffinal^2 = FT-init / FT-final 500^2 /ffinal^2 = 1.5Vg / 1.6Vg ffinal = 516 Hz

Ch 18.4 ”sis”ter and swing

You are pushing your little sister on a swing. The center of mass of your sister to the pivot point is 1.6 meters. If your little sister wants to go the highest, how often should you push?

ω² = g / L

(2π/T)² = 9.8/1.6

T = 2.54 seconds

Ch 18.5 #45 An air column in a glass tube is open at one end and closed at the other by a movable piston. The air in the tube is warmed above room temperature, and a 384-Hz tuning fork is held at the open end. Resonance is heard when the piston is 22.8 cm from the open end and again when it is 68.3 cm from the open end. (a) What speed of sound is implied by these data? (b) How far from the open end will the piston be when the next resonance is heard?
For resonance in a narrow tube open at one end, f = nv / 4L (n = 1, 3, 5…)	(a) For n = 1 and n = 3			(b) f(n=1) = nv / 4L 384 = 5(350) / 4L L = 1.14 m
	f(n=1) = nv / 4L 384 = 1(v) / 4(0.228) v = 350 m/s	f(n=3) = nv / 4L 384 = 3(v) / 4(0.683) v = 350 m/s

Ch 18.6 #49

An aluminum rod 1.60 m long is held at its center. It is stroked with a rosin-coated cloth to set up a longitudinal vibration. The speed of sound in a thin rod of aluminum is 5100 m/s. (a) What is the fundamental frequency of the waves established in the rod? (b) What harmonics are set up in the rod held in this manner? (c) What If? What would be the fundamental frequency if the rod were copper, in which the speed of sound is 3,560 m/s?

(a) v = f λ v = f 2L 5100 = f 3.2 f = 1590 Hz

A-N-A L = ½λ Rod clamped at center; the center will be a node and each ends will be antinodes

(b) The center must always be a node and the ends, antinodes; so harmonics are odd. 1^st: ANA 3^rd: ANANANA (skips 2^nd: ANANA) 5^th: ANANANANANA (skips 4^th: ANANANA).

c) v = f λ v = f 2L 3560 = f 3.2 f = 1110 Hz

Ch 18.7 guitar

You’ve decided to enhance your guitar. You only use 3 of the 5 strings, so you are going to replace the remaining two with the same type of string of the same length (which requires minimal modification). One of the strings will have the frequency of 1000 Hz and a tension of 300 N. What would lower the tension in the second string if you desire a beat frequency of 10 Hz?

f^2 λ^2 = FT / μ f^2 = FT / μλ^2	finit^2 = FT-init / μλ^2 ffinal^2 = FT-final / μλ^2	finit^2 / ffinal^2 = FT-init/FT-final 1000^2 / 990^2 = 300 / FT-final FT-final = 294 N
Note: 1. μ constant since same string 2. λ is constant since identical boundary conditions (same length)