When a mass of 5 kg is suspended from a spring of negligible mass and spring constant K, it oscillates with a periodic time `2pi`. If the mass is removed, the length of the spring will decrease by

Options

A. `(g)/(K)` metre
B. g metre
C. `2pi ` metre
D. `(m)/(K)` metre

Answer

B

Explanation

For the oscillating spring ` T=2pi sqrt((m)/(K))`
where m is the mass suspended from the spring. If x is the extension produced in the spring, then mg=Kx
`therefore` Spring constant `K=(mg)/(x)`
`therefore T=2pi sqrt((m)/((mg)/(x)))=2pi sqrt((x)/(g))`
But `T=2pi therefore 2pi =2pi sqrt((x)/(g)) therefore x=g`
When the mass m is removed, the spring will have its original unstretched length i.e. its length decreases by x.
`therefore` Decrease in length =x=g metre