Determination of the spring constant and effective mass of a given spiral spring and hence to calculate the rigidity modulus of the material of the spring.

Determination of the spring constant and effective mass of a given spiral spring and hence to calculate the rigidity modulus of the material of the spring.


Firstly, we measure the diameter of the spring several times using slide calipers and take the mean value. Then we attach it to a stand. Diameter of the wire is then measured by screw gauge. Then twice the diameter of wire is substracted from outer diameter of spring to get inner diameter and average of that two is taken as diameter of spring.[1] Now a certain mass is hung from spring. Mean position is marked. We lift it and let go causing a vertical oscillation. We measure the time for 50 oscillation and record the corresponding data. The same process is repeated for 5 times. Then graph plotting (T^2 vs M) ,calculation of k and and error analysis is done based on the obtained data.


Spring constant is a  characteristic of a spring which is defined as the ratio of the force affecting the spring to the displacement caused by it. Usually denoted with the letter "k" in formulae, as in the formula F =k − x , where "F" is the force applied and "x" is the displacement. Hooke's Law states that the restoring force of a spring is directly proportional to a small displacement.[2] In equation form, we write,

Where x is the size of the displacement. The proportionality constant k is specific for each spring. The object of this virtual lab is to determine the spring constant k.[3]  

The effective mass of the spring in a spring-mass system when using an ideal spring of uniform linear density is 1/3 of the mass of the spring and is independent of the direction of the spring-mass system (i.e. horizontal, vertical, and oblique systems all have the same effective mass). This is because external acceleration does not affect the period of motion around the equilibrium point. 
We can find the effective mass of the spring by finding its kinetic energy. This requires adding all the mass elements' kinetic energy, and requires the following integral, where u is the velocity of mass element : 

Modulus of Rigidity, G: Modulus of rigidity of a spring is also known as the spring constant (Shear Modulus) is the coefficient of elasticity for a shearing force. It is defined as "the ratio of shear stress to the displacement per unit sample length (shear strain)".Modulus of Rigidity can be experimentally determined from the slope of a stress-strain curve created during tensile tests conducted on a sample of the material. The definition of Modulus of Rigidity: the ratio of shear stress to the displacement per unit sample length (shear strain).[5] 


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 m0g 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.[6] If k is the spring constant of the spring and the extension of the spring, then restoring force = - k
Let the spring is in equilibrium with mass mo attached as in Fig. 02 (a), and  so  we can write  
     m0g = kl
=> l=g/k mo         … … … … … … … … … … … … …(1)    
k is the spring constant and g is the acceleration due to gravity. 
Equation (1) is an equation of straight line slope of this line is given by
Slope =  g/k,
K = g/slope               … … … … … … … … … … … … …(2) 
We can plot l vs mo 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 we know from our earlier schools if the mass of the spring were negligible then the period of oscillation.
That would be given by,

Due to the mass, m of the spring an extra term m′ will be added with the mass of the load mo in the above mentioned equation. So, the period of oscillation is,
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, 
m' = m/3 … … …. …. …. … … … … … … … … … … … (4) 

So, from equation (3), we get,
For different mass, mo of the load we find different periods of oscillation, T. If we draw a graph by plotting m0 along X axis and corresponding T2 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. T2 = 0 there. We can find the X coordinate of the point, (i.e. the value of mo at that point) by putting T2 = 0 in the above mentioned equation,              

That means x coordinate of the point is equal to the negative value of the effective mass. So, if we draw a T2 vs. mo graph, it will be a straight line and its x-interception gives us the effective mass of the spring.
If  n  is the rigidity modulus of the material of the spring than it can also prove that, 

N = Number of turns in the spring
R = Radius of the spring 
K = Spring constant 
r = Radius of the wire of the spring 


1. A spiral spring
2.Convenient masses with hanging arrangement
3.Clamp or a hook attached to a rigid framework of heavy metal rods
4. Weighting balance
5. Stop watch and 
6. Scale 

Experimental Data

No. of observation
Loads mo in gm
Extension in cm
No. of vibrations
Total time  in sec
Period T in sec




Spring constant, K = Mg = 7229.508  dynes/cm
From graph (2) effective mass = 40 gm
a) No. of turns N in the spring = 162
b) Radius of the spring,   R = (D+d)/4 = (2.785+2.575)/4  = 1.34 cm
c)  Radius of the wire of the spring, r = 0.65975 cm

Percentage error


The spring constant of a given spiral spring is 7229.508  dynes/cm and effective mass is 40 gm. And the rigidity modulus of the material of the spring is  8.8439 × 10^11 dynes/cm  with an error of 6.5% .


Measurement of radius of wire/spring and weight-

1. Some instrumental error may arise if main scale zero doesn’t coincide with circular scale     zero. That have been taken care of by adding or substracting the error.
2. If the wire is not uniform then readings have to take at different points and mean of them is   the best reading.
3. While hanging the weights that should lie along the axis of spring. This doesn’t occur during experiment.

Measurement of time period -

1.  Firstly measurement of time period is not accurate due to personal observation error.
2.  Oscillations should occur in vertical plane but it oscillates little in a horizontal plane also.


In the experiment, there are some important apparatus. These apparatus are very useful to find out the exact value of spring material. Spring constant can be measured accurately by following the precaution of these apparatus. There will be small amount of error if the value is taken accurately. 


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