Oscillation of a spring

Introduction

When the position of a particle / mass varies periodically between two points or about a certain point then it is said to be in oscillatory motion. Restoring force is required for the oscillation of a particle or mass. Whenever, a particle or mass is forced to change its stable position, then in the presence of restoring force the particle or mass starts oscillating between two points or about a central point. Whenever the spring is either stretched or compressed, the restoring force comes into existence. We can compress or stretch the spring either horizontally or vertically. Due to the restoring force in the spring, the spring starts oscillating between two points or about a central point.  In this article, first we will discuss horizontal oscillation and then vertical oscillation along with the combination of springs.

Hooke’s Law and Spring Constant

Hooke’s law says that “the restoring force developed in the spring due to extension / compression , is directly proportional to the extension / compression of the spring. It is generally written as, “F = -k x”. “F” represents the restoring force, “’x” is the displacement of spring either in the form of extension or compression and “k” is constant of proportionality. As we have negative signs in Hooke’s law, we can say that the direction of restoring force and the direction of displacement have opposite directions. 

The constant of proportionality is known as stiffness constant /spring constant. Stiffness of the spring is indicated by  this Spring Constant   i.e. higher value of “k” means more stiffness and lower value of spring constant means less stiffness of the given spring. 

Horizontal Oscillation of a spring 

We will assume a system in which we have a block of mass m attached to a massless spring with stiffness constant or force constant or spring constant of the spring as “k”. We will place the spring mass system on a frictionless / smooth horizontal surface.

 Let x0 be the equilibrium position or mean position of mass m at time t = 0 and the spring is in relaxed condition i.e. mass is at rest. If we pull the mass horizontally through a small displacement x towards right from its equilibrium position and then after releasing, it will start back and forth oscillating about this mean / equilibrium position. If the restoring force in the spring is assumed to be “F” then,

F= -K x (from Hooke’s law)

Now, 

md2xdt2= -K x ( from Newton’s second law)

Now, comparing above equation with simple harmonic motion, we have

2= km 

So , ω=km

Frequency of oscillation can be given as f= 2π= 12πkm

Time period of oscillation, T= 1f= 2πmk

Vertical Oscillation of a spring

We will consider a spring-mass system. In this system a block of mass m has been attached at one end of  a mass-less spring and the spring has “k” as spring constant. We will suspend the spring mass system from a rigid support and assume that no air drag is present. In this case, the force constant is given by,

k= mgl

Now, if we pull down the block by a small distance “x”, a restoring force “-kx” acts vertically upward and pulls the block in upward direction. Due to this restoring force, the block will return to its initial position and it will continue to move in upward direction. It will overshoot the initial equilibrium condition and the spring will be compressed by distance “’y” in upward direction. In this situation, the restoring force will be in a downward direction. Due to this, the block will again move in downward direction and overshoot the equilibrium point. The system will continue to execute oscillations in vertical direction.

Now, comparing above equation with simple harmonic motion, we have

2= km 

So , ω=km

Frequency of oscillation can be given as f= 2π= 12πkm

Time period of oscillation, T= 1f= 2πmk = 2πmlmg = 2πlg

Parallel and Series Combination of springs

If we have two massless springs with spring constant “k1” and “k2” connected in parallel to the same block of mass “m” at one end. Then equivalent spring constant of this combination is given as:

keq = k1 + k2 .

Time period of oscillation, T= 1f= 2πmk1 + k2

If we have two massless springs with spring constant k1 and k2 connected in series to a block of mass “m”. Then equivalent spring constant of this combination is given as

k= k1*k2k1+k2.

Time period of oscillation, T= 1f= 2πm(k1+k2)k1* k2

Conclusion

In the above article, we discussed the oscillations of a spring in horizontal and parallel directions. We observed that spring constant is a very important factor in oscillation of a spring as it decides the stiffness of the spring. If we have higher value of spring constant, then we have greater stiffness of the given spring, which means greater restoring for a particular displacement of the spring and for lower value of stiffness of the spring, lesser restoring force will be developed in the spring for a particular displacement of the spring. 

The spring constant of the spring affects the time period of oscillation of the spring mass system as the time period of oscillation and the square root of the spring constant are inversely related to each other which means that for higher value of spring constant, time period has lesser value and for lower value of the spring constant we get higher value for time period “T” of oscillation.