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Introduction To Spring Constant

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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 Constant

Spring-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.

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