A string secured at its ends is stretched with a force of tension `F`. The mass of the string and the force of gravity can be neglected. A point weight with a mass m is attached to the middle of the string. The time period of small oscillations of the weight is `2pi sqrt((ml)/(xF))`. the value of `x` is ________
(Neglect any change in the magnitude of tension of the string during oscillations)

Answer

4

Explanation

The force acting on the weight delected from the position of equilibrium is `2F sin phi` Since the angle `phi` is small, it may be assumed that
Force `=(4Fx)/(l) or F = kx`, where `k = (4F)/(l)`. By using the formula `T =2pi sqrt((m)/(k))`
We get the following expression for the sough quantity: `T = 2pi sqrt((ml)/(4F))`