A 0.1 kg mass is suspended from a wire of negligible mass. The length of the wire is 1 m and its cross sectional area is 4.9 times 10^-7 m^2. IF the mass is pulled a little in the vertically downward direction and released, it performs simple harmonic motion of angular frequency 140 rad.s^-1. If the young's modulus of the material of the wire is n times 10^9 N.m^-2 Calculate the value of n.

A 0.1 kg mass is suspended from a wire of negligible mass. The length

1.	A 0.1 kg mass is suspended from a wire of negligible mass. The length of the wire is 1 m and its cross sectional area is 4.9 times 10^-7 m^2. IF the mass is pulled a little in the vertically downward direction and released, it performs simple harmonic motion of angular frequency 140 rad.s^-1. If the young's modulus of the material of the wire is n times 10^9 N.m^-2 Calculate the value of n.
Answer» Solution :YOUNG modules of the material of the WIRE, `Y=(F/A)/((DeltaA)/L)=(FL)/(A DeltaL) or,F=((YA)/L)DeltaL`......(1) IF mass m is pulled by a length `DeltaL` then restoring FORCE developed in the wire, `F=k DeltaL`…….(2) Comparing equation (1) and (2) we GET , `k=(YA)/L` Angular frequency , `omega=sqrt(K/m)=sqrt((YA)/(ML))` or,`140=sqrt((ntimes10^9times4.9times10^-7)/(0.1times1))=70sqrtn` `therefore sqrtn=2 or n=4`