1. You are studying guitar strings and are able to measure a few of (but not all

Question

1. You are studying guitar strings and are able to measure a few of (but not all) the standing wave frequencies for a particular string. If you measure frequencies of 360 Hz, 600 Hz, 840 Hz, and 1080Hz, what is the highest possible value for the fundamental frequency?

2.A steel piano string for the note A is about 0.55 m long and is under a tension of 590 N. If the fundamental frequency is 440 Hz, what is the string's diameter? (Assume steel has a density of 7,800 kg/m3.) (answer is in mm)

3.Two neighbors communicate with a homemade string phone (see figure below). The simple phone consists of two ends of a string (total mass of 19 g) attached to the bottom of two paper cups. The string is stretched between the two homes at a tension of 21 N over the distance of 36 m.

(a) How long does it take communication to travel on the string from one home to the other?

(b) How long would it take to communicate by yelling out the window?

(c) How long would it take to communicate over cell phones if the nearest cell tower is 5.3 km away? (Consider only the propagation time of the radio waves and assume that the call is already connected.)

Explanation / Answer

1) n*fo = 360

(n+1)*fo = 600

fo = 600-360 = 240 Hz

2) f = (1/2L)*sqrt(T/u)

u = T/(4L^2*f^2 = (590)/(4*0.55^2*440^2) = 0.0025
u = M/L = rho*A*L/L = rho*A = 0.0025

A = 0.0025/rho

pi*d^2/4 = A

d = 641.355e?6 m = 0.641355 mm

3)

a) t = L/V = L*sqrt(T/u) = L/sqrt(TL/m)

t = 36/sqrt((21*36)/0.019) = 0.18 s

b) t = L/Vs = 36/343 = 0.105s

c) t = L/c = 36/(3*10^8) = 120e?9 s

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