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Introduction To Spring Constant

In the last article, we learnt the basic concepts of SHM. Now we shall try to visualise the spring-mass system. Before that, we will see what spring constant is. Force by the action of the spring is given by,

\(\begin{array}{l}F\end{array} \)
=
\(\begin{array}{l}-kx\end{array} \)

k is known as the spring constant or stiffness constant.

Unit of spring constant is N/m.

There are different types of spring. For example torsion spring which works due to turning of the spring.

We can also visualise this spring-mass motion with the help of uniform circular motion.

Spring Constant

Spring-mass motion using uniform circular motion

Suppose at

\(\begin{array}{l} t\end{array} \)
=
\(\begin{array}{l} 0 \end{array} \)
the particle is as shown in the figure. If we take the projection of that on the x-axis it is at origin. At
\(\begin{array}{l}t \end{array} \)
=
\(\begin{array}{l} T_1\end{array} \)
, if we take the projection along x-axis it is in positive x-direction and having positive velocity. Similarly, at
\(\begin{array}{l}t\end{array} \)
=
\(\begin{array}{l}T_2\end{array} \)
, the projection is in negative x-axis and having negative velocity. By observing this we can say we take the projection of uniform circular motion on the x-axis it represents an SHM. Sometimes this approach is very useful in solving the numerical problem as compared to equation-based approach.

How to find spring constant when a number of springs are connected together? Suppose we have a situation as shown below:

Spring ConstantSpring-mass system 1

Suppose the mass is displaced by x towards right, so the force exerted by S2

\(\begin{array}{l}F_2\end{array} \)
=
\(\begin{array}{l}F_2=-k_2x\end{array} \)
(Towards left)

Force exerted by

\(\begin{array}{l}S_1\end{array} \)
,

\(\begin{array}{l}F_1\end{array} \)
=
\(\begin{array}{l}-k_1x\end{array} \)
(Towards left)

So the total restoring force

\(\begin{array}{l}(F)\end{array} \)
=
\(\begin{array}{l} F_1 + F_2\end{array} \)

\(\begin{array}{l}F\end{array} \)
=
\(\begin{array}{l}- (k_1 + k_2) x\end{array} \)

Hence, equivalent

\(\begin{array}{l}k \end{array} \)
=
\(\begin{array}{l}k_1 + k_2\end{array} \)

Using a similar approach try to find equivalent spring constant for the following?

Spring Constant

Spring-mass system 2

Also, the time period for the spring mass system can be shown to be,

\(\begin{array}{l}T\end{array} \)
=
\(\begin{array}{l}2π\sqrt{\frac{m}{k}}\end{array} \)

Where k is the equivalent stiffness constant.

The relationship between the spring constant and the displacement is explained in the video with the help of a spring balance.

Stay tuned with BYJU’S to learn more about spring constant, SHM and much more.

Frequently Asked Questions – FAQs

What is the unit of the spring constant?

The unit of the spring constant is N/m.

What is a spring?

Spring is an object that can be deformed by force and then return to its original shape when the force is removed.

What is the formula to find the force by the action of the spring?

The force by the action of the spring is given by the formula: F=-kx

How is the spring classified based on the displacement due to force applied?

  • Constant Force Spring
  • Variable Rate Spring
  • Linear Spring
  • What is the formula to find the spring force?

    The Spring force formula is given by: F = k(x – x0)

    Test your knowledge on Introduction To Spring Constant

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