Oscillations of a spring – restoring force

Oscillation can be defined as a movement back and forth about the equilibrium point. Generally, the word vibration is used when talking about mechanical oscillation. The soundwaves reaching our eardrums and our heartbeat also create oscillation. A simple example of oscillation is a pendulum, a spring, and a guitar string. A spring moves downwards and upwards, which creates oscillatory movement. Oscillations can be divided into two types- damped oscillation and undamped oscillation. Alternating current or AC waves is an example of undamped oscillation.

Sampled oscillation can also be divided into three types, namely, underdamped oscillation (damping constant < 1), overdamped oscillations (sampling constant > 1), and critically damped oscillation (damping constant = 1). Some examples of damped oscillation are amplitude of a pendulum and vibrations of a tuning fork.

Some Examples of Oscillation

We will now discuss some of the most common examples of oscillation before moving on to the oscillations of a spring. 

A simple pendulum is an example of oscillation. The pendulum of the clock moves to and fro, which creates an oscillatory motion. Vibrations are mechanical oscillations. When an object vibrates, it means that the object moves back and forth between two points about the equilibrium or mean or central point. 

Similarly, the movement of spring is also an example of oscillation.

Oscillations of a Spring

Imagine a spring with no force applied to it. At this stage, the spring is at a resting position. It is in an equilibrium position. At this position, the spring-mass system is in its natural position. Now, imagine that an external force is applied to the spring. Then, a force will direct the spring towards its equilibrium position. This force is known as the restoring force. Now again, imagine an internal force applied to it. Again, restoring force will direct the spring towards the equilibrium position. Thus, if spring is displaced from its equilibrium position, then a restoring force will direct it towards equilibrium.

Spring oscillation equation can be written as:

Ts​ =2π √m/k

In this spring oscillation equation, Ts denotes the time period of the spring; m is used to denote the mass of the spring; and k is known as the spring constant. This equation basically means that the time period of the spring mass oscillator is directly proportional to the square root of the mass of the spring, and it is inversely proportional to the square of the spring constant.

Let us now look at the horizontal and vertical oscillations of the spring.

Horizontal oscillations of a spring

Imagine a mass m is attached to a spring on one end, and the other end is fixed to a wall, and the spring is placed on a smooth surface. When no force is applied to the spring, it remains at its resting or equilibrium position. However, if the spring is displaced from its mean or equilibrium position outwards and then the force is withdrawn, the spring will return to its resting position in an inward direction. Thus, we can notice that the restoring force always acts in the opposite direction to the displacement. The restoring force also has a proportional relationship with the displacement.

We know that restoring force is:

F = -kx.

Newton’s second law states that:

F = ma.

Thus, we can write, ma =-kx.

If we compare it with the simple harmonic motion equation a = −ω2x, we get:

ω2 = k/m.

Now, ω = √( k/m ).

However, T = 2π / ω.

Thus, the time period T = 2π √(m/k).

Thus the Frequency will be f = 1/T = 1/2 π √( k/m ).

Let us now have a look at the vertical spring oscillation.

Vertical spring oscillation

Imagine two springs are attached together vertically, and a mass is placed on one end of the spring, and the other end is attached to a support, such as a wall. Now, let the displaced distance be y, which is equal to the displacement of first spring y1 and displacement of second spring y2. Let k1 and k2 be the force constants. We will get the following:

F = −k1 y1, and 

F = −k2 y2.

F in these equations is the restoring force. 

The total displacement will be 

y = y1 + y2

−F [1/k1 + 1/k2].

And, F = −ky

y= – F/k.

Thus, from the above-mentioned equation, we get:

-F/k = −F [1/k1 + 1/k2];

 T (time period) = 2π . √(m(k1+k2)/k1k2);

f (frequency) = 1/2 π . √(k1k2 / m(k1+k2) ).

If both the springs have the same spring constant,

k1 = k2 = k;

f= 1/2 π . √(k/2m).

Students often assume that the spring mass oscillator period is dependent on the amplitude. However, this is not true. When the amplitude increases, it simply means that mass is traveling more distance to complete one cycle. But the increase in amplitude increases the restoring force. This increase in the restoring force thereby increases the acceleration of mass proportionally such that the mass covers more distance at the same time. Therefore, the increase in amplitude has zero net effect on the oscillation period.

​Conclusion

So far we have covered the oscillation of a spring. Oscillation occurs when a body moves to and fro about an equilibrium point at a fixed interval of time. Movement of a pendulum and a spring can be defined as an oscillatory movement. Oscillation of a spring is its displacement from the equilibrium position. Spring oscillation equation can be mathematically expressed as Ts =2π √(m/k). We have also discussed types of oscillations namely vertical oscillation of spring and horizontal oscillation of a spring.