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

Options

A. `2pisqrt((m)/(K))`
B. `2pisqrt((m(YA+KL))/(YAK))`
C. `2pisqrt((mYA)/(KL))`
D. `2pisqrt((mL)/(YA))`

Answer

B

Explanation

Let us consider the wire also as a spring. Then the case becomes two springs attached in series. The equilvalent spring constant is:
therefore `(l)/(K_(eq))=(l)/(K)+(l)/(K')`
where K' is the spring constant constant of the wire
`K_(eq)=(KK')/(K+K')`
Now, `Y=(F//A)/(DeltaL//L)=(F)/(A)xx(L)/(DeltaL)` `(F)/(DeltaL)=(YA)/(L)=K'`
we know that time period of the system
`T=2pisqrt((m)/(K_(eq)))=2pi sqrt((m(K+K'))/(KK'))` implies `T=2pisqrt((m)/(K)[(K+YA//L)/(YA//L)])=2pisqrt((m(KL+YA))/(KYA))`