To determine the spring constant of a spring by measuring its stretch versus applied force, to determine the spring constant of a spring by measuring the period of oscillation for different masses, and also to investigate the dependence of period of oscillation on the value of mass and amplitude of motion.
If the applied force (mass) is to remain the same while the vertical displacement is increased, there period will remain the same. However, if the vertical displacement is held constant while the applied force in increased, the period will increase. This is in accordance with the derived equation T = 2π (M / k)0.5.
See attached sheet.
k = 3.53 N/m
|A (m)||∆t1 (s)||∆t2 (s)||∆t3 (s)||∆t avg. (s)||αt (s)||T (s)|
|M (kg)||∆t1 (s)||∆t2 (s)||∆t3 (s)||∆t avg. (s)||αt (s)||T (s)||T2 (s2)|
ms = 0.00940 kg
Slope = 10.5 s2/kg
Y-Intercept = 0.0687 N
k = 3.75 N/m
C = 0.694
% difference = 6.04%
1. Do the data from Part 1 verify Hooke’s Law? State clearly the evidence for your answer.
The data correlate close to Hooke’s Law, but not quite. The law states that F = -ky, where F is in this case Mg and y equals the negative displacement. After graphing forces versus displacement, a value of 3.53 N/m was determined as the spring constant. However, when applying this value to the equation and using recorded displacement values, the calculated force come up less than the actual for used. For example, in the first trial y = -0.055 m. Multiplying that value by the extrapolated spring constant gives a theoretical force of 0.194 N, but the actual force used was 0.244 N. All other trials yield a similar lowball theoretical force.
2. How is the period T expected to depend upon the amplitude A? Do your data confirm this expectation?
Period is not expected to depend upon amplitude, as suggested by the equation T = 2π (M / k)0.5, where amplitude is absent as a variable. The data confirms this expectation, as the period was nearly the same for each trial.
3. Consider the value you obtained for C. If you were to express C as a whole number fraction, which of the following would best fit your data (1/2, 1/3, 1/4, 1/5)?
The obtained value of C is 0.694, which is closest to 1/3.
4. Calculate T predicted by the equation T = 2π (M / k)0.5 for M = 0.050 kg. Calculate T predicted by T = 2π ( [M + Cms] / k )0.5 for M = 0.050 kg and your value of C. What is the percent difference between them? Repeat for a value for M of 1.000 kg. Is there a difference in the percent differences? If so, which is greater and why?
T predicted using the first equation and M = 0.050 kg is 0.726 s. T predicted using the second equation and M = 0.050 kg is 0.771 s. This is a percent difference of 6.01%.
T predicted using the first equation and M = 1.000 kg is 3.24 s. T predicted using the second equation and M = 1.000 kg is 3.26 s. This is a percent difference of only 0.615%. The greater percent difference occurs at the lower weight because the weight of the spring is almost insignificant at higher weight. The proportion of the mass to the spring is so great that it has almost no effect on the calculation.
During part one of the experiment, the vertical displacement of a spring was measured as a function of force applied to it. The starting position of the spring was recorded using a stretch indicator. Mass was added to the spring, and the displacement was recorded. This was repeated with various amounts of mass. From these data, a graph of force versus displacement was plotted, and a linear fit slope revealed the spring constant. In this endeavor, the spring constant was valued at 3.53 N/m.
However, when applying this spring constant to the recorded displacements in the Hooke’s Law equation, the calculated forces are lower than the recorded forces. In similar manner, when rearranging Hooke’s Law to solve for displacement, the calculated displacements are larger than the actual recorded displacements. This means there was human error, most likely in terms of not being precise with the displacement readings because the recordings for the masses used were accurate. Because such small masses were used, any error in displacement readings was augmented. The spring used may also not have been perfect.
During part two of the experiment, the period of the spring was measured as amplitude changed while mass remained constant. The period remained nearly the same throughout every trial, which was to be expected. Any differences in period may be accounted to inadequate stopwatch usage and inaccurate starting displacements throughout the trials. It should be noted that at amplitude of 0.100 m, the hook lost contact with the spring for a split second at the apex of oscillation, which accounts for its oddity in period. This was something that could not be avoided; at that amplitude the spring pulled the hook up too quickly which caused the loss of contact. The resulted graph of period versus amplitude yielded a linear fit slope of close to 0 (-0.247 s/m), which was predicted.
During part three of the experiment, the period of the spring was measured as mass was varied while amplitude remained constant. As the mass was increased, the period also increased. This was not surprising considering the given equations. The square of the period versus mass for each trial was plotted and a linear fit was taken. The slope and y-intercept of this line was then used to determine the spring constant and C, the fraction of the spring’s mass that should be taken into account for the equation T = 2π ( [M + Cms] / k )0.5. The calculated value of k was 3.75 N/m, which is only 6.04% different from the value determined earlier of 3.53 N/m. The value of C was determined to be 0.694, which is closest to the whole number fraction of 1/3. Any error during parts two and three can be attributed to inaccurate stopwatch recordings and slight variance in displacement and release of the masses at each amplitude.
T = 2π (M / k)0.5
T = 2π ( [M + Cms] / k )0.5
slope = 4π2 / k
y-intercept = 4π2Cms / k