Explain the horizontal oscillations of a spring.

Explanation

`(i)` Let `'m'` be the mass of a block attached to a mass less spring
`(ii) 'k'` be the stiffness constant or force constant or spring constant of the spring.
`(iii) 'x_(0)'` be the mean or equilibrium position of the block.
`(iv) 'x'` be the small displacement of the block towards right, and the released.
`(v)` The block will oscillate back and forth about `x_(0)`
`(vi)` If `F` be the restoring force, then it is proportional to the `'x'`



`F prop x`
`F=-kx`
`m(d^(2)x)/(dt^(2))=-kx`
`(d^(2)x)/(dt^(2))=-(k)/(m)x`
`omega^(2)=(k)/(m)`
`omega=sqrt((k)/(m))rads^(-1)`
`f=(omega)/(2pi)=(1)/(2pi)hertz`
`T=(1)/(f)=2pisqrt((m)/(k))` second