GR9677 #93

bchase
Posts: 4
Joined: Wed Jan 05, 2005 1:15 pm

GR9677 #93

Postby bchase » Sun Oct 23, 2005 1:19 pm

Anyone have any ideas for the following GR9677 #93?

A particle of mass m moves in potential
V(x) = 1/2 k x^2 (x < 0)
V(x) = mgx (x > 0)

For a given energy E, what is the period of motion?

danty
Posts: 19
Joined: Fri Sep 30, 2005 6:40 pm

Postby danty » Mon Oct 24, 2005 12:25 pm

The period of oscillation will be the sum of the half period of the particle moving in the parabolic potential plus the half period of the particle moving in the V(x)=mgx potential.

You will probably know that the period of oscillation in a parabolic potential is given by the equation T=2pi*(m/k)^1/2. So the half period is pi*(m/k)^1/2. By looking at the possible answers you can see that the only answer containing the above term is D. Therefore D is the correct answer.

The above is a quick way to find the solution to this problem with out calculating the period of the motion.

However, if you want to calculate the half period of the motion in the V=mgx potential, you must first find the extremum value of x, derived from E-V(x)=0, or xmax=E/mg, and then find the time required for a displacement from xmax, to x=0 , which is derived from the equation xmax=(1/2)gt^2 , or t=(2E/mg^2)^1/2 (the acceleration is constant and it is g). Think now that in a full period the particle makes this move 4 times , so the half period is 2 times the amount t. So the half period is
2(2E/mg^2)^1/2 .

Therefore the period in this composite potential is:

pi*(m/k)^1/2 + 2(2E/mg^2)^1/2

bchase
Posts: 4
Joined: Wed Jan 05, 2005 1:15 pm

Postby bchase » Mon Oct 24, 2005 2:02 pm

Thank you!




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