# Physics: Analytic Geometry and Mass

Principle of Physics (PHY111)

Lab

Experiment no: 4

Name of the experiment: Determination of the spring constant and effective mass of a given spiral spring

Theory

In this experiment a spring is suspended vertically from a clamp attached to a rigid frame work of heavy metal rods. At the bottom end (which is the free end) of the spring a load of mass, m0 is suspended. So the force acting on the spring is the weight m0 g of the load which acts vertically downward and the spring gets extended. Due to the elastic property of the spring, it tries to regain its initial size, hence applies a counter force on the load, which is called the restoring force of the spring. According to Hooke’s law, magnitude of this restoring force is directly proportional to the extension of the and the direction of this restoring force is always towards the equilibrium position. If k is the spring constant of the spring and l is the extension of the spring, then restoring force = - k l

Figure 1: a) A vertically suspended spring, b) A load of mass m0 is attached at the bottom end of the spring and c) The spring is oscillating, the load is y distance above the equilibrium position. Let the spring is in equilibrium with mass m0 attached as in figure 1 (a), and so we can write 0

= klgm => m0 k

g

l = (1) Here k is the spring constant and g is the acceleration due to gravity. Equation (1) is and equation of straight line slope of this line is given by- Slope =

k

g

=> k =

slope

g

(2) We can plot l vs. m0 graph and determine its slope to determine k. If the load is slightly pulled down and released, the spring will oscillate simple harmonically. Suppose, at time t the velocity of the load is v and the spring is compressed by a distance y above the point C. As you know from your earlier schools if the mass of the spring were negligible then the period of oscillation would be given by

k

m

T

0

= 2π

Due to the mass, m of the spring an extra term m′ will be added with the mass of the load m0 in the above mentioned equation. So, the period of oscillation is,

k

mm

T

+ ′

=

0

2π (3) m′ is called to be the effective mass of the spring. It can be showed that m′ is related with the mass of the spring by following equation:

3

m

m′ = (4) Please see appendix A (provided in the soft copy of this script in the server) to learn how equations (4) and (3) come.

From equation (3) we get,

m

k

m

k

T += ′

2

0

2

2

44 ππ

For different mass, m0 of the load we find different periods of oscillation, T. If we draw a graph by plotting m0 along X axis and corresponding T

2

along Y axis, it will be a straight line. The point where the line intersects the X axis, its y-coordinate is 0, i.e., T

2

=0 there. We can find the X coordinate of the point, (i.e. the value of m0 at that point) by putting T

2

=0 in the above mentioned equation.

m

k

m

k

+= ′

2

0

2

44

0

ππ=> −= mm ′

0

That means x coordinate of the point is equal to the negative value of the effective mass. So, if we draw a T

2

vs. mo graph, it will be a straight line and its...

Please join StudyMode to read the full document