Question

A spring with a force constant of 111Nm is attached to the ceiling. When an unknown mass is attached it pulls the spring down 0.340meters to its equilibrium position. The mass is then pulled 9cm below its equilibrium position and oscillates back and forth. Determine the period of oscillation (in seconds)?

Answer 1

T = 1.17 s

Step-by-step explanation:

As we know that unknown mass is attached with a spring

then in the equilibrium position the force of gravity on the mass is balanced by the spring force on the block

so we have

`mg = kx`

now we have

`(m)/(k) = (x)/(g)`

here we know that

`x = 0.340 m`

`g = 9.81 m/s^2`

`(m)/(k) = (0.340)/(9.81)`

`(m)/(k) = 0.035`

now we know that time period of the SHM is given as

`T = 2\pi \sqrt{(m)/(k)}`

`T = 2\pi √(0.035)`

`T = 1.17 s`