Oscillations Of A Spring-mass System

An oscillation is a movement about the equilibrium point that goes back and forth. Generally speaking, mechanical oscillation is referred to as vibration. Our heartbeat and the sound waves that reach our eardrums both cause oscillation. A guitar string, a spring, and a pendulum make up a basic oscillation demonstration. 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.

JEE Main 2025: Physics Formula | Study Materials | High Scoring Topics | Preparation Guide

JEE Main 2025: Syllabus | Sample Papers | Mock Tests | PYQs | Study Plan 100 Days

NEET 2025: Syllabus | High Scoring Topics | PYQs

This Story also Contains

1. What is Spring Force?
2. The time period of the Spring mass system:
3. Key points:
4. Combinations of Spring
5. 1. Series combination of spring
6. 2. Parallel combination of spring
7. Solved Examples Based on Spring System
8. Summary
9. Frequently Asked Questions (FAQs):Q1: How Does Mass Affect the Period of a Spring?

In this article, we will cover the 'Oscillations of a spring-mass system’ concept. This concept is part of the chapter Oscillations and Waves, which is a crucial chapter in Class 11 physics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), National Eligibility Entrance Test (NEET), and other entrance exams such as SRMJEE, BITSAT, WBJEE, VITEEE and more. Over the last ten years of the JEE Main exam (from 2013 to 2023), fourteen questions have been asked on this concept. And for NEET one question was asked from this concept.

What is Spring Force?

Spring force is also called restoring force. F= -kx where k is the spring constant and its unit is N/m and x is net elongation or compression in the spring. Spring constant (k) is a measure of stiffness or softness of the spring

Here, the -ve sign is because the force exerted by the spring is always in the opposite direction to the displacement.

The time period of the Spring mass system:

• Oscillation of a spring in a verticle plane finding time period of spring using the force method.

Let $x_0$ be the deformation in the spring in equilibrium. Then $k x_0=m g$. When the block is further displaced by x, the net restoring force is given by $F=-\left[k\left(x+x_0\right)-m g\right]$ as shown in the below figure.

Using $F=-\left[k\left(x+x_0\right)-m g\right]$ and $k x_0=m g$.

We get $F=-k x$

comparing it with the equation of SHM i.e $F=-m \omega^2 x$

we get,

$ \omega^2=\frac{k}{m} \Rightarrow T=2 \pi \sqrt{\frac{m}{k}}$

similarly

$ \text { Frequency }=n=\frac{1}{2 \pi} \sqrt{\frac{k}{m}}$

• Oscillation of spring in the horizontal plane

For the above figure, Using the force method we get a Time period of spring as

$
T=2 \pi \sqrt{\frac{m}{k}}
$
and
Frequency $=n=\frac{1}{2 \pi} \sqrt{\frac{k}{m}}$

Key points:

1. The time period of a spring-mass system depends on the mass suspended

$ T \propto \sqrt{m} \quad \text { or } \quad n \propto \frac{1}{\sqrt{m}}$

2. The time period of a spring-mass system depends on the force constant k of the spring

$ T \propto \frac{1}{\sqrt{k}} \quad \text { or } \quad n \propto \sqrt{k}$

3. The time period of a spring-mass system is independent of acceleration due to gravity.

4. The spring constant k is inversely proportional to the spring length.

$ \text { As } \quad k \propto \frac{1}{\text { Extension }} \propto \frac{1}{\text { Length of spring }(l)}$

$\text { i.e } k l=\text { constant }$

That means if the length of the spring is halved then its force constant becomes double.

5. When a spring of length l is cut in two pieces of length l₁ and l₂ such that $ l_1=n l_2$

So using

$ \begin{aligned}
& l_1+l_2=l \\
& n l_2+l_2=l \\
& (n+1) l_2=l \Rightarrow l_2=\frac{l}{n+1} \\
& \text { similarly } l_1=n l_2 \Rightarrow l_1=\frac{l * n}{(n+1)}
\end{aligned}$

If the constant of a spring is k then

using $k l=$ constant
i.e $k_1 l_1=k_2 l_2=k l$

we get,

Spring constant of first part $k_1=\frac{k(n+1)}{n}$

Spring constant of second part $k_2=(n+1) k$

And ratio of spring constant $\frac{k_1}{k_2}=\frac{1}{n}$

6. If the spring has a mass M and mass m is suspended from it, then its effective mass is given by $m_{e f f}=m+\frac{M}{3}$

And

$T=2 \pi \sqrt{\frac{m_{e f f}}{k}}$

Recommended Topic Video

Combinations of Spring

There are two types of combinations of spring

1. Series combination of spring

If 2 springs of different force constants are connected in series as shown in the below figure

then k=equivalent force constant is given by

$\frac{1}{K_{e q}}=\frac{1}{K}=\frac{1}{K_1}+\frac{1}{K_2}$

Where are spring constants of spring 1 & 2 respectively.

Where $K_1$ and $K_2$ are spring constants of spring $1 \& 2$ respectively.
Similarly, If $\mathrm{n}$ springs of different force constants are connected in series having force constant $k_1, k_2, k_3 \ldots \ldots$ respectively

then

$\frac{1}{k_{e f f}}=\frac{1}{k_1}+\frac{1}{k_2}+\frac{1}{k_3}+\ldots \ldots \ldots$

$\text { If all the } \mathrm{n} \text { springs have the same spring constant as } K_1 \text { then } \frac{1}{k_{e f f}}=\frac{n}{k_1}$

2. Parallel combination of spring

If 2 springs of different force constants are connected in parallel as shown in the below figure

then k= equivalent force constant is given by

$K_{e q}=K=K_1+K_2$

$\text { where } K_1 \text { and } K_2 \text { are spring constants of spring } 1 \& 2 \text { respectively. }$

Similarly, If $\mathrm{n}$ springs of different force constants are connected in parallel having force constant $k_1, k_2, k_3 \ldots \ldots$ respectively then $K_{e q}=K_1+K_2+K_3 \ldots$

$\text { If all the } \mathrm{n} \text { spring have the same spring constant as } K_1 \text { then } K_{e q}=n K_1$

Solved Examples Based on Spring System

Example 1: A mass M is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes SHM of time period T. If the mass is increased by m, the time period becomes 5T/3. Then the ratio of m/M is :

1) $3 / 5$
2) $25 / 9$
3) $16 / 9$
4) $5 / 3$

Solution:

Initially, $T=2 \pi \sqrt{M / k}$
Finally, $\frac{5 T}{3}=2 \pi \sqrt{\frac{M+m}{k}}$
$
\begin{aligned}
& \therefore \quad \frac{5}{3} \times 2 \pi \sqrt{\frac{M}{k}}=2 \pi \sqrt{\frac{M+m}{k}} \\
& \text { or } \frac{25}{9} \frac{M}{k}=\frac{M+m}{k} \\
& \text { or } 9 m+9 M=25 M \\
& \text { or } \frac{m}{M}=\frac{16}{9}
\end{aligned}
$

Hence, the answer is option (3).

Example 2: A spring having a spring constant 'k' is loaded with a mass 'm'. The spring is cut into two equal parts and one of these is loaded again with the same mass. The new spring constant is :

1) $-2K$
2) $k$
3) $2 k$
4) $k^2$

Solution:

$\text { Spring Constant }(k) \alpha \frac{1}{\text { length of the spring }(l)}$

$\text { as length becomes half, } k \text { becomes twice i.e } 2 k$

Hence, the answer is the option (3).

Example 3: A particle of mass 200 grams executes SHM. The restoring force is provided by a spring of force constant 80 N/m. The period ( in seconds) of oscillation is :

1) 0.31

2) 0.15

3) 0.05

4) 0.02

Solution:

Time period of oscillation for a spring-mass system

$
T=2 \pi \sqrt{\frac{m}{k}}
$
wherein
$m=$ mass of block
$\mathrm{k}=$ spring constant
$
T=2 \pi \sqrt{\frac{m}{k}}=2 \pi \sqrt{\frac{0.2}{80}}=0.31 \mathrm{~s}
$

Hence, the answer is the option (1).

Example 4: A spring has a certain mass suspended from it and its period for vertical oscillation is T. The spring is now cut into equal halves and the same mass is suspended from one of the halves. The period of the vertical oscillation is now :

1) $T / 2$
2) $\frac{T}{\sqrt{2}}$
3) $\sqrt{2} \mathrm{~T}$
4) $2 T$

Solution:

Period of oscillation for the spring-mass system

$
T=2 \pi \sqrt{\frac{m}{k}}
$

Also
$
\text { Spring constant }(k) \alpha \frac{1}{\text { Length }(l)}
$

When the spring is half in length, then $\mathrm{k}$ becomes twice
$
\therefore T^{\prime}=2 \pi \sqrt{\frac{m}{2 k}} \Rightarrow \frac{T^{\prime}}{T}=\frac{1}{\sqrt{2}} \Rightarrow T^{\prime}=\frac{T}{\sqrt{2}}
$

Hence, the answer is the option (2).

Example 5: Two bodies of masses 1 kg and 4 kg are connected to a vertical spring, as shown in the figure. The smaller mass executes simple harmonic motion of angular frequency 25 rad/s, and amplitude 1.6 cm while the bigger mass remains stationary on the ground. The maximum force (in Newtons) exerted by the system on the floor is (take g = 10 ms^-2).

1) 60

2) 10

3) 20

4) 40

As we know

$\begin{aligned}
& T=2 \pi \sqrt{\frac{m}{k}} \\
& \text { or } \frac{2 \pi}{w}=2 \pi \sqrt{\frac{m}{k}} \text { or } \frac{1}{25}=\sqrt{\frac{1}{k}} \\
& m=1 \mathrm{~kg}, w=25 \mathrm{rad} / \mathrm{s} \\
& \text { or } k=625 \mathrm{Nm}^{-1}
\end{aligned}$

Now, as the upper block is performing SHM. So the force exerted by the system on the floor is maximum when Spring is compressed by the maximum amount after the initial equilibrium condition.

Maximum compression is $A=1.6 \mathrm{~cm}$

So for the figure below

$
a_{\max }=\omega^2 A=625 \times 1.6 \times 10^{-2}
$

So applying Newton's 2nd law on the upper block we get
$
\begin{aligned}
& f-m g=m a_{\max }=m \omega^2 A=1 \times 625 \times \frac{1.6}{100}=10 N \\
& \Rightarrow f=20 N
\end{aligned}
$

Now by applying Newton's 2nd law on the lower block, we get
$
N_P=M g+f=(4 \times 10)+20=60 N
$

So the maximum force exerted by the system on the floor is 60 N.

Hence, the answer is option (1).

Summary

The amplitude affects the spring-mass oscillator period. This is untrue, though. Simply put, a rise in amplitude indicates that mass is moving a greater distance in a single cycle. However, the restoring force increases as the amplitude increases. As a result of this rise in the restoring force, the mass accelerates proportionately faster and travels a greater distance simultaneously. As a result, the oscillation period is unaffected by the amplitude increase.

Frequently Asked Questions (FAQs):
Q1: How Does Mass Affect the Period of a Spring?

Ans: 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.

Q 2: Does Frequency Depend on Mass?

Ans: Yes, frequency depends on mass.

Q 3: What is Hooke's Law?

Ans: Hooke's Law states that the force exerted by a spring is proportional to its displacement from the equilibrium position and is given by F = -kx, where k is the spring constant and x is the displacement.

Q 4: How is the angular frequency related to the spring-mass system?

Ans: The angular frequency ω is related to the period and is given by: $\omega=\sqrt{\frac{k}{m}}$