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JEE Main 2026 Preparation: Question Papers, Solutions, Mock Tests & Strategy Unacademy » JEE Study Material » Physics » Oscillations of a spring

Oscillations of a spring

In this article oscillations of a spring, we will discuss oscillation of a spring, it's equation, horizontal and vertical spring oscillation Conditions at Mean Position, and the Amplitude in Oscillation motion.

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Introduction

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 die.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 with 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/k1];

T (time period) = 2π . √(m(k1+k2)/k1k1);

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).

Conclusion 

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 travelling 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 the 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.

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