Question

2. In lab, you attach one end of an ideal spring (of negligible mass) to a wall and the other end of the spring to a 0.200kg Pascar that sits at rest on a frictionless, horizontal track. You then pull the Pascar horizontally and let go, causing it start it oscillating. The elapsed time from when the Pascar first moves through the equilibrium point to the second time it moves through the equilibrium point is 2.60s. Find the spring constant, k, of the spring.

Answer 1

The time taken by the object to move from equilibrium point to maximum amplitude and then back to the equilibrium point is the half time period of the oscillation given as 2.60s.

So Time period of this motion(T) = 2*(half time period)

T = 2*2.6s = 5.2s
We know that T = (2*pi)/w

where "w" is the omega

and "omega" is the angular frequency.
w = (2*pi) /(5.2) = 1.21 rad/sec

We know that for a spring mass system

w = (k/m)^0.5

where k is the spring constant and m is the mass of the object.

So squaring both sides and rearranging

k = m.w²

k = 0.2.(1.21)² N m^-1

k = 0.292 N m^-1