Question

PROBLEM 14.2: "If an object on a horizontal frictionless surface is attached to a spring, displaced, and then released, it will oscillate. The object is displaced a distance 0.120 m from its equilibrium position and released with zero initial speed. Then after a time 0.800 s, its displacement is found to be a distance 0.120 m on the opposite side, and it has passed the equilibrium position once during this interval. a) Find the amplitude. b) Find the period. c) Find the frequency."

Answer 1

a)
`A = 0.120\,m`, b)
`T = 1.6\,s`, c)
`f = 0.625\,hz`

Step-by-step explanation:

The known conditions of the motion are:

`x(0) = 0.120\,m`

`x(0.8) = - 0.120\,m`

a) The amplitude is the half of the difference between both positions:

`A = (x(0)-x(0.8))/(2)`

`A = (0.120\,m-(-0.120\,m))/(2)`

`A = 0.120\,m`

b) The period of oscillation is equal to the double of the time between both positions:

`T = 2\cdot \Delta t`

`T = 2\cdot (0.8\,s)`

`T = 1.6\,s`

c) The frequency of the oscillation is:

`f = (1)/(T)`

`f = (1)/(1.6\,s)`

`f = 0.625\,hz`