Hooke's Law and Simple Harmonic Motion

Springs are modeled using Hooke's Law:

F = - k Dx,

where Dx is the deviation from the equilibrium position, and k is the spring constant (measured in g / s2).

In addition, the oscillatory nature of the spring depends on the spring constant and the mass of the object bouncing on the end of the spring:

w2 = k / m,

where w is the "angular frequency" of the oscillation, and equals 2p divided by the time for one oscillation (its "period"). Note that the dimensions of k over m are one over time squared.

We will measure the spring constants of two different springs in two different ways, using the following equipment: a stand, 2 springs, a weight pan, weights, a stop watch and a balance.

Name:

Lab Partners:

When entering numeric data, use exponentials: ie., 1.6 * 10-19 = 1.6E-19.

Procedure

  1. Describe the first spring (ie., looser or tighter):

    Measure the elongation of the first spring Dx1, m for weights m from 50 to 200 grams in 10 gram increments:

    Dx1, 50 = cm Dx1, 130 = cm
    Dx1, 60 = cm Dx1, 140 = cm
    Dx1, 70 = cm Dx1, 150 = cm
    Dx1, 80 = cm Dx1, 160 = cm
    Dx1, 90 = cm Dx1, 170 = cm
    Dx1, 100 = cm Dx1, 180 = cm
    Dx1, 110 = cm Dx1, 190 = cm
    Dx1, 120 = cm Dx1, 200 = cm
    Be sure the weights are still when you measure the distance. Also, be sure that the pointer is at zero before you begin adding weights (some springs won't zero on this apparatus, since they are too long; all elongations are relative to where the pointer was when the weight pan was empty). All masses are relative to zero (for this part of the experiment, pretend the pan is massless. Why can we do that?)

  2. Describe the second spring:

    Repeat the last step with the second spring:

    Dx2, 50 = cm Dx2, 130 = cm
    Dx2, 60 = cm Dx2, 140 = cm
    Dx2, 70 = cm Dx2, 150 = cm
    Dx2, 80 = cm Dx2, 160 = cm
    Dx2, 90 = cm Dx2, 170 = cm
    Dx2, 100 = cm Dx2, 180 = cm
    Dx2, 110 = cm Dx2, 190 = cm
    Dx2, 120 = cm Dx2, 200 = cm

  3. Repeat the last step with both springs (one connected to the other):

    Dx3, 50 = cm Dx3, 130 = cm
    Dx3, 60 = cm Dx3, 140 = cm
    Dx3, 70 = cm Dx3, 150 = cm
    Dx3, 80 = cm Dx3, 160 = cm
    Dx3, 90 = cm Dx3, 170 = cm
    Dx3, 100 = cm Dx3, 180 = cm
    Dx3, 110 = cm Dx3, 190 = cm
    Dx3, 120 = cm Dx3, 200 = cm

  4. Place a 100 g mass on the first spring. Pull the mass down slightly from its equilibrium position and release it, simultaneously starting the stop watch. Record the time for 30 oscillations (30 trips where the mass returns to the point at which you let it go):

    t1 = s

  5. Repeat the last step with the second spring:

    t2 = s

  6. Measure the mass of the weight pan:

    mp = g

Analysis

  1. Perform a least squares fit on each of the data sets for weight (y = mass times g) versus elongation (Dx) ("i" is the spring number; g = 981 cm / s2):

    Xi = ( Dxi, 50 + ... + Dxi, 200 ) / 16

    X1 = cm

    X2 = cm

    X3 = cm

    Y = ( 50 * g + ... + 200 * g ) / 16

    Y = dynes

    mi = ( 16 * Xi * Y - ( 50 * g * Dxi, 50 ) - ... - ( 200 * g * Dxi, 200 ) ) / ( 16 * Xi2 - Dxi, 502 - ... - Dxi, 2002 )

    m1 = g / s2

    m2 = g / s2

    m3 = g / s2

    bi = Y - mi * Xi

    b1 = dynes

    b2 = dynes

    b3 = dynes

    Graph your data with the least squares fit lines (all three graphs on the same page).

    What can you conclude?

  2. For each spring, compute the angular frequency wi by dividing 2 p by the time for one oscillation (its period):

    wi = 2 p / ( ti / 30 )

    w1 = Hz

    w2 = Hz

    (1 Hz = 1 / s.)
    Multiply the mass of the weight PLUS the weight pan (why the pan?) by the square of w for each spring

    ki = ( 100 + mp ) * wi2

    k1 = g / s2

    k2 = g / s2

    and compare with the slopes from 1. What can you conclude?

©2004, Kenneth R. Koehler. All Rights Reserved. This document may be freely reproduced provided that this copyright notice is included.

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