Experiment Name: Determination of the Spring constant and effective mass and hence to calculate and rigidity modulus of the material of the spring.
Objectives:
Determination of the spring constant and effective mass of a wire.
Calculating the rigidity modulus of the material of the spring.
Learning Outcomes:
Understanding the concept of simple harmonic motion.
Students will be able to understand the variation of extension of spring at different loaded mass.
Understanding the concept of Hook’s law and restoring force.
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 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 restoring force is directly proportional to the extension of the spring 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 = -kl
Let, the spring is in equilibrium with mass m attached as in figure and so we can write
m g = k Ɩ
k = mlg --------------- (i)
Here k is the spring constant and g is the acceleration due to gravity.
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,
T=2mo+ m'k--------- --(iii)
m' is called the effective mass of the spring. It can be showed that m′ is related with the mass of the spring by following equation,
m'= m3
From equation (iv)→
T2=42km0+42k m΄
∴0=42km0+42k m΄
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 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 m0 at that point) by putting T2 = 0 in the above mentioned equation.
m0 = - m′
That means x coordinate of the point is equal to the negative value of the effective mass. So, if we draw a T2 vs. m0 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. Then it can also be proved that
η = 4NkR3r4------------------------------------ (iv)
Where N = number of turns in the spring, R = radius of the spring and r = radius of the wire of the spring and k = spring constant.
Apparatus: A spiral spring, convenient masses with hanging arrangement, a hook attached to a rigid framework of heavy metal rods, weighing balance, stop watch and scale
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