One end of a long metallic wire of length (L) is tied to the ceiling. The other end is tied to a massless spring of spring constant . (K.A) mass (m) hangs freely from the free end of the spring. The area of cross- section and the Young's modulus of the wire are (A) and (Y) respectively. If the mass is slightly pulled down and released, it will oscillate with a time period (T) equal to :

Explanation

Let the mass m be pulled downwards through distance y. Let the extension in wire of length L be `y_(2)` and in spring be `y_(1)`, figure, then,
`y=y_(1)+y_(2)` …(i)
Since there is no mass between spring and wire, the restoring force in the spring and wire will be the same say F.
For a spring, according to Hooke's law,
`F=ky_(1) or y_(1)=F//k`
For wire, by definitino of Young's modulus,
`y=(FL)/(Ay_(2)) or y_(2)=(FL)/(AY)`
Putting values in (i), we get
`y=(F)/(k)+(FY)/(AY)=F[(AY+kL)/(AYk)]`
or `F=[AYk)/(AY+kL)]y`
As `Fprop y` and it is directed towards the mean position beign restoring force, hence when mass is left free after pulling will execute linear SHM
Here, spring factor `=(Yk)/(AY+kL)` and
inertia factor `=` mass of body `=` m
`:. T=2pisqrt((i n ertia fact o r)/(spri ng fact o r))`
`=2pisqrt((m(AY+kL))/(AYk))`