A metal wire of diameter `1 mm` is held on two knife edges by a distance `50 cm`. The tension in the wire is `100 N`. The wire vibrating with its fundamental frequency and a vibrating tuning fork together produce `5 beats//s`. The tension in the wire is then reduced to `81 N`. When the two are excited, beats are heard at the same rate. Calculate
(a) frequency of a fork and
(b) the density of material of wire.

Explanation

Let the frequency oftuning fork be n, then in the first case the fundamental frequencyofthe wirewill be(n + 5), whichis given as
`(n + 5) = (1)/(2l) sqrt(((T)/(mu)))`
Here T = 100 N, l = 50 cm = 0.5m , on substituting values, we get
`(n+ 5) = (1)/(2 xx 0.5) sqrt(((100)/(mu))) = (10)/(sqrt(mu))` In the second case, T = 81 N. In this case the frequency of wire will be (n -5) Thus we have
`(n - 5) = (1)/(2xx 0.5) sqrt(((81)/(mu))) = (9)/(sqrt(mu))`
Subtractingequation-(6.253)from equation-(6.252), we get
`10 = (10)/(sqrt(mu)) - (9)/(sqrt(mu)) = (1)/(sqrt(mu))`
`mu = (1)/(100)` = 0.001 kg/m
But we know that linear mass density'ofa wire ofcross sectional radius r and density p is given as
`mu = pir^(2) P`
`P = (m)/(pir^(2)) = (0.01)/(3.14(5 xx10^(-4))^(2))`
= `12732.5 kg//m^(3)`
From equation-(6.252), we have
`n + 5 = (10)/(sqrt(0.01))` = 100 Hz
n = 100 - 5
= 95 Hz