A vertical spring of constant k = 54.43 N/m is attached to a surface on its bott
ID: 1542660 • Letter: A
Question
A vertical spring of constant k = 54.43 N/m is attached to a surface on its bottom side. If nothing is attached to the spring its equilibrium position is x0 = 0. A 1-kg-mass (m1) is then fixed to the top of the spring and it compresses a distance d1 to a new equilibrium position x1 = 0. (a) What is the value of d1? Now an identical mass (m2) is placed on top of m1 and the two are pushed down compressing the spring an additional distance d = 1 m from the equilibrium posi- tion x1 = 0, as shown in the figure. The two are released from rest and m2 is launched up in the air, leaving contact with m1 at x1 = 0. (b) What is the speed of the two masses as they pass x1 = 0 for the first time? (hint: make all of your distances in the spring potential energy function rela- tive to x0 = 0) What is the (c) angular frequency and (d) maximum displacement (from x1 = 0) of the oscillation of m1 after m2 is no longer in contact? (e) Plot the position as function of time for m1, while m2 is not in contact. Do this for two complete cycles and use a grid to make the graph clear. Use the provided scaling and set t = 0 s as the moment m2 loses contact with m1. (f) Either analytically, or numerically using the graph from part (e), determine the time and po- sition that m2 first comes back into contact with m1.
Explanation / Answer
a)spring constant k= 54.43N/m
if m-0 x=0.
when m1=1kg,,
k= mg/d2
d2=mg/k
d2= 1* 9.8 /54.43
d2= 0.18005m
d2=18.005 cm