Mr. Burtness Name______

Lab

Simple Harmonic Motion - Bouncing mass on a Spring

The velocity of an object in Simple Harmonic Motion as a function of position is:

v=+/-vmaxÖ(1-x2/A2), where x is the displacement from the equilibrium position, and A is the amplitude of oscillation, xmax ( ½ peak to peak distance). This comes from conservation of energy principles:

Uspring = ½ kx2 (k = spring constant from Hooke’s Law: F=-kx). K.E. = ½ mv2

Since mechanical energy is conserved: ½ mv2 + ½ kx2 = ½ kA2 + ½ m[v(=0)]2 = ½ mvmax2

  1. Solve this underlined equation for: v= ___ P.E.max = K.E.max, so ½ kA2 = ½ mvmax2
  2. Solve this for k/m=, and substitute into equation #1.


Laboratory Experiment:

·  Hang a mass from a spring at least 70 cm above a rangefinder. Note the stretch a 500g mass gives the spring to determine the spring constant, k.

·  Use the range finder to collect data (ball-bounce setting on the CBR) on the bouncing of the mass. The data should look very smooth.

  1. Record A, vmax , and T, the period of oscillation.
  1. Calculate vmax =2pA/T and compare to your experimental value.
  1. Substitute vmax /A=Ö(k/m) into the above equation and solve for T. This is the equation for the period of motion.
  1. To see if your data agrees with the equation (#2) above, you can calculate v from the position graph as follows. First, shift the entire position graph down so it is symmetric with the x-axis. Do this by noting a time where velocity has reached a maximum. Next read your position graph at this time, xmax . QUIT the RANGER program, and “ENTER” L2-xmax STOà L2. This will re-center your position graph.

On your TI calculator: L2 = position, L3 = velocity, L2=x (displacement from equil.)

v=vmax{abs[1-(L2)2/A2].5} àL5.


  1. Graph L1 vs. L3, and L1 vs. L5. This will show you how well they agree. (Note: absolute value is required in the calculation to prevent negative square roots that would result from experimental data.) An actual comparison of these velocities is shown below.
  2. To see that the motion is sinusoidal, set Y1=vmaxsin{(k/m).5x} You may have to shift the curve left or right to get complete agreement, but the shape will be the same as the above picture.

SHMCBR.doc