Question

Two identical 4.70g wires, each 1.90m long and supporting a 114kg weight, hang next to each other.

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

(delta)M = ? kg

Answer 1

linear mass density, mue = m/L

= 4.7*10^-3/1.9

= 2.47*10^-3 kg/m

let f is the initial frequency

f = v/lamda

f = sqrt(T/mue)/(2*L)

= sqrt(M*g/mue)/(2*L)

let f' is the frequency when deltaM mass is added

f' = sqrt((M+deltaM)*g/mue)/(2*L)

f(beat) = f' - f

5 = sqrt((M+deltaM)*g/mue)/(2*L) -sqrt(M*g/mue)/(2*L)

5*2*L = sqrt((M+deltaM)*g/mue) -sqrt(M*g/mue)

5*2*1.9 = sqrt((114+deltaM)*9.8/2.47*10^-3) - sqrt(114*9.8/2.47*10^-3)

19 = 63*sqrt(114+deltaM) - 672.54

sqrt(114+deltaM) = (19+672.54)/63

sqrt(114+deltaM) = 10.97

114+deltaM = 120.49

deltaM = 6.5 kg <<<<<<<<<<<-----------------Answer