Question

Two identical 4.70 g wires, each 1.80 m long and supporting a 106 kg weight, hang next to each other.

Part A

How much mass should you add to one of these weights so that, in their fundamental harmonics, these wires produce a 5.00 Hz beat? Assume that the wires are strong enough that they do not stretch with the added weight.

ΔM =

  kg  

Answer 1

tension string T = M*g

M = mass attached

speed of wave in string = sqrt(T/u)

u = linear mass density = m/L

m = mass of wire

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

f = (1/2L)*sqrt(MgL/m)

f2 = (1/2L)*sqrt((M+dM)*g*L/m)

f1 = (1/2L)*sqrt((M*g*L/m) = (1/(2*1.8))*sqrt(106*9.8*1.8/0.0047) = 175.2

beats = f2 - f1 = 5

f2 - 175.2 = 5

f2 = 180.2 Hz

(1/2L)*sqrt((M+dM)*g*L/m) = 180.2

(1/(2*1.8))*sqrt((106+dM)*9.8*1.8/0.0047) = 180.2

dM = 6.14 kg <<<======ANSWER