A point mass m is suspended at the end of a massless wire of length l and cross section. If Y is the Young's modulus for the wire, obtain the frequency of oscillation for the simple harmonic motion along the vertical line.

Explanation

The origin length of wire is l
The downwards force acting on the block is mg. If T is the tension in the wire, then
`T = mg` because this in eqilibrium
According to Hooke's law.
`y = ("Long stress")/("Long strain") = (T//A)/(Delta l//l)`
where Delta l is the length of the wire. Therefore .
`T = YA (Deltal)/(l)`
Putting `T = mg`, we get `mg = YA (Deltal//l)` (i)
Let the block be displaced down a little through a distance x during oscillators. The tension in the wire acting upwards will be `(YA)/(l) (Deltal + x) = (mg)/(Delta l) (Delta l + x)`
The downward force is mg
Therefore, resultant downward force on the mass will be
`F = - [(mg)/(Delta l) (Delta l + x) - mg]`
`= - ((mg)/(Delta l)) x`
`= - kx`
Thus, the net force is directly proportional to the displace, meat, but oppositely directed . Hence the motion of the block is SHm
Comparing with standerd equation of SHM, we gwt
`F = - m omega^(2) x`
We get `omega = sqrt((k//m))`
`:. T = (2 pi)/(omega) = 2 pi sqrt(m//k) = 2 pi sqrt(ml//AY)`
and frequency of vibration `= n = (1)/(T) = (1)/(2 pi) sqrt((AY)/(ml))`