Courses for Kids
Free study material
Offline Centres

Spring-Mass System

Last updated date: 30th Nov 2023
Total views: 21.6k
Views today: 1.21k
hightlight icon
highlight icon
highlight icon
share icon
copy icon

Introduction to Spring-Mass System

A spring-mass system consists of a block attached to a free end of the spring.

Let’s say a block of mass m, is attached to a spring, as shown below:

[Image will be Uploaded Soon]

When it is displaced by x, the spring expands and then comes back to its original position.

[Image will be Uploaded Soon]

In the absence of any resistance, this spring oscillates infinitely. These motions are simple harmonic.

So, the spring-mass system helps us find out the period of any object performing the simple harmonic motion. 

Lets’ understand the steps to find the period of an SHM:

Steps to Find a Period of any SHM

Step 1: Find the Equilibrium Position

An equilibrium position is a position where FNET = 0.

Now, let’s find the position where the net force is zero.

Here, we will write a balanced force equation.

Step 2: Displace the Object by x 

Displace the mass slightly (either to the right or to the left) from its equilibrium position. On finding a new FNET, we mathematically, arrive at FNET = - kx.

Where k = force constant (or spring constant).

Step 3: Find the Period from the Formula 

Since k = mω² ⇒ ω² = \[\frac{m}{k}\] ⇒ ω = \[\sqrt{\frac{m}{k}}\] 

We know that T = \[\frac{2 \pi}{\omega}\] =  \[\frac{2 \pi}{\sqrt{\frac{m}{k}}}\], we get

⇒ T = 2π\[\sqrt{\frac{m}{k}}\] 

This is the period of an SHM.

Now, let’s find out the period of a spring-mass system.

Period of Mass-Spring System

Let’s displace mass m by a distance x:

Step 1: Initially, the mass is at rest, when displaced by a force F, it reaches to x0, as shown below:

[Image will be Uploaded Soon]                       

So, the equation for the equilibrium position is F = kx0. At this position, the net force on the system is zero.

Step 2: Now, let’s displace it further by x from its equilibrium position, then F = kx.

The force indicated backward by the green arrow in fig.a is k(x0 + x). 

However, the forward force is still F.

We can see that backward force > forward force. Then, the net force is:

FNET  = - k(x0 + x) + F 

= F - k(x0 + x) = F - kx0 - kx 

∵ F =  kx0

= F - F - kx 


FNET  = - kx

This is the required mass-spring system equation.

So, the period of mass-spring system is:

T = 2π\[\sqrt{\frac{m}{k}}\] 

This means heavier is the mass, lesser is the period of spring. The period of spring is independent of the constant force.

We know that frequency f = 1/T, so putting the value of T in this equation, we get:

f  = \[\frac{1}{2 \pi}\]\[\sqrt{\frac{k}{m}}\] 

This is the natural frequency of the spring-mass system (also known as the resonance frequency of a string).

Now, let’s find the differential of the spring-mass system equation.

Spring-Mass System Differential Equation

Let’s consider a vertical spring-mass system:

A body of mass m is pulled by a force F, which is equal to mg. The displacement on the spring is x0. So, the force acting on it will be F = - kx0.   

[Image will be Uploaded Soon]                        

Now, the body is displaced further from its equilibrium by a distance x. 

Then, the restoring force is - k(x0 + x), which is greater than mg.

So,   - k(x0 + x) + mg ⇒ mg - k(x0 + x)

Here, kx0 = mg

⇒ mg - mg - kx

 FNET  = - kx….(1)

This spring accelerates by a, which is equal to \[\frac{d^{2}x}{dt^{2}}\] 

We know that F = ma = m\[\frac{d^{2}x}{dt^{2}}\]…(2)

Here, the net force acting on the mass = motion of the mass

Now, (1) = (2), we get

- kx = m d²x/dt²

m\[\frac{d^{2}x}{dt^{2}}\] + kx = 0

This is the required spring-mass system differential equation.                

Now, let’s consider a two mass-spring system.

Two Mass-Spring System

Two mass-spring systems can be arranged in two ways. They are:

  1. Series combination of springs

  2. The parallel combination of springs

Let’s discuss these one-by-one:

1. Series Combination of Springs

[Image will be Uploaded Soon]

Generally, on pulling string systems, it displaces by x0. Then, on pulling it from the equilibrium position, it displaces by a distance x.

However, in two springs systems, the displacement on each spring is different.

So, the displacement at spring S1 and S2 are x1 and x2, respectively.

Then, the net displacement is:

x = x1 + x2….(1)

The forces on each spring are:

F1 = - k1x1, and

 F2 = - k2x2

So, x1 = - \[\frac{F}{k1}\], and  x2 = - \[\frac{F}{k2}\]

Now, putting the value of in eq (1)

 x = (-\[\frac{F}{k1}\]) + (-\[\frac{F}{k2}\])

We know that F = - kx


- \[\frac{F}{k}\] = (-\[\frac{F}{k1}\]) + (-\[\frac{F}{k2}\])

So, we get

\[\frac{1}{k}\] = \[\frac{1}{k1}\] + \[\frac{1}{k2}\]  ⇒ k = \[\frac{k1k2}{k1+k2}\] 

We know that T = 2π\[\sqrt{\frac{m}{k}}\] 

So, for a series combination, T is:


T = 2π\[\sqrt{\frac{m(k1+k2)}{k1k2}}\] 

2. The Parallel Combination of Springs

[Image will be Uploaded Soon]

The net force acting on the springs arranged in parallel is:

F = F1 + F2

So, - kx = - k1x1 - k2x2

We get

k = k1 + k2

We know that T = 2π\[\sqrt{\frac{m}{k}}\]

So, for a parallel combination, T is:

T = 2π\[\sqrt{\frac{m}{k1+k2}}\]

FAQs on Spring-Mass System

Q1: How Does Mass Affect the Period of a Spring?

Ans: By formula, 

T = 2π√(m/k) 

We can say that the time period is directly proportional to the mass of the body attached to the spring. This means if a heavier object like a truck is attached to it, it oscillates slowly.

Q2: A Spring of Force Constant 1400 is Mounted on a Horizontal Table. A Mass of 4 kg is Attached to the Free End of the String, Displaced by 5 m is Released, then What is the Frequency of Oscillation of the Mass?

Ans: Here, k = 1400, m = 4 kg, x  = 5 m, f = ?

Now, putting these values in the formula, 

 f = 1/2π √(k/m) = 1/(2 x 3.14) √(1400/4)

On solving, we get,

f  = 2.97 s⁻¹

Q3: A Spring of Force Constant 1500 is Mounted on a Horizontal Table. A Mass of 5 kg is Attached to the Free End of the String, is Displaced by 6 cm, and Released. Find the Maximum Acceleration of the Mass.

Ans: Here, k = 1500, m = 5 kg, x  = 6 cm, a = ?

 We know a = ω²y

Acceleration will be maximum when a = y, then

AMAX = ka/m = (1500 x 0.06)/5 = 18 ms².

Q4: Does Frequency Depend on Mass?

Ans: Yes.

By the relation,  f = 1/2π √(k/m) 

We can say that frequency depends on the force constant, k, and the mass m.