please help me in my lab report i need to find those equations Provide a general
ID: 1428426 • Letter: P
Question
please help me in my lab report i need to find those equations
Provide a general formula for an elastic force (The Hooke's law). Solve this relation for a static spring constant. Express the static spring constant as a function of SLOPE1. Provide numerical substitutions and a final result with units.
Error propagation for the static spring constant. To find the error in the static spring constant take a derivative of the static spring constant as a function of SLOPE1. Slope 1 is a variable in this case. Provide numerical substitutions in this formula. Provide units for a final result.
Explanation / Answer
if an object applies a force to a spring, then the spring applies an equal and opposite force to the object. Hooke’s law gives the force a spring exerts on an object attached to it with the following equation:
F = –kx
where the minus sign shows that this force is in the opposite direction of the force that’s stretching or compressing the spring. (k is called the spring constant, which measures how stiff and strong the spring is. x is the distance the spring is stretched or compressed away from its equilibrium or rest position.)
The restoring force, F, of a stretched spring is proportional to its elongation, x, if the deformation is not too great. This relationship for elastic behavior is known as Hooke's law and is described by
F = -kx (eq. 1),
where k is the constant of proportionality called the spring constant. The spring’s restoring force acts in the opposite direction to its elongation, denoted by the negative sign. For a system such as shown in figure 1, the spring's elongation, x – x0, is dependent upon the spring constant, k, and the weight of a mass, mg, that hangs on the spring. If the system of forces is in equilibrium (i.e., it has no relative acceleration), then the sum of the forces down (the weight) is equal and opposite to the sum of the forces acting upward (the restoring force of the spring), or
m g = k (x – x0) (eq. 2).
Comparing equation 2 with the form for the equation of a straight line (y = mx + b), we can see that if we plot the force produced by different masses (mg) as a function of the displacement from equilibrium (x-x0), the data should be linear and the slope of the line will be equal to the spring constant, k, whose standard metric units are N/m. If the mass is pulled so that the spring is stretched beyond its equilibrium (resting) position, the restoring force of the spring will cause an acceleration back toward the equilibrium position of the spring, and the mass will oscillate in simple harmonic motion. The period of vibration, T, is defined as the amount of time it takes for one complete oscillation, and for the system described above is
(eq. 3),
where me is the equivalent mass of the system, that is, the sum of the mass, m, which hangs from the spring and the spring's equivalent mass, me-spring, or
me = m + me-spring (eq. 4)
Note that me-spring is not the actual mass of the spring, but is the equivalent mass of the spring. It is not the actual mass because not all of the mass pulls down to act in concert with the weight pulling down. Its theoretical value for our system should be approximately 1/3 of the actual mass of the spring. Substituting equation 4 into equation 3 and squaring both sides of the equation yields:
Expanding the equation:
(eq. 5)
Therefore, if we perform an experiment in which the mass hanging at the end of the spring (the independent variable) is varied and measure the period squared (T2 ; the dependent variable), we can plot the data and fit it linearly. Comparing equation 5 to the equation for a straight line (y = mx + b), we see that the slope and y-intercept, respectively, of the linear fit is:
and y-intercept =
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