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.

Options

A. `(1)/(2pi)sqrt((YA)/(mL))`
B. `2pi sqrt((mL)/(YA))`
C. `(1)/(pi) sqrt((YA)/(mL))`
D. `pi sqrt((mL)/(YA))`

Answer

A

Explanation

Frequency depends on spring factor and inertia factor. In this case, stress `= (mg)/(A) = (I)/(L)`
(where, I is extension) Now, Young's modulus Y is given by
`Y = ("stress")/("strain")=(mg//A)/(I//L)or mg = (YAI)/(L)`
So, `KI = (YAI)/(L)" "(becuase mg = kI)" "("k is force constant")`
`therefore" "k = (YA)/(L)`
Now, frequency is given by, `n = (1)/(2 pi) sqrt((k)/(m))`
`= (1)/(2 pi) sqrt(((YA)/(mL)))`