# 1.7 Block – spring system in shm  (Page 2/2)

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$⇒{F}_{\text{net}}=ma=-ky$

$⇒a=-\frac{k}{my}$

This relation on comparison with SHM equation “ $a=-{\omega }^{2}x$ ” yields same set of periodic expressions as in the case of horizontal block-spring arrangement :

$⇒\omega =\sqrt{\left(\frac{k}{m}\right)}\phantom{\rule{1em}{0ex}}⇒T=2\pi \sqrt{\left(\frac{m}{k}\right)}\phantom{\rule{1em}{0ex}}⇒\nu =\frac{1}{2\pi }\sqrt{\left(\frac{k}{m}\right)}$

In this case, however, we can obtain alternative expressions as well for the periodic attributes as spring force at equilibrium position is equal to the weight of the block,

$mg=-k{y}_{0}$

Dropping negative sign and rearranging,

$⇒\frac{m}{k}=\frac{{y}_{0}}{g}$

Hence, the alternative expressions of periodic attributes are :

$⇒\omega =\sqrt{\left(\frac{g}{{y}_{0}}\right)}\phantom{\rule{1em}{0ex}}⇒T=2\pi \sqrt{\left(\frac{{y}_{0}}{g}\right)}\phantom{\rule{1em}{0ex}}⇒\nu =\frac{1}{2\pi }\sqrt{\left(\frac{g}{{y}_{0}}\right)}$

Clearly, the extension of the spring owing to the weight of the block in vertical orientation has no impact on the periodic attributes of the SHM. One important difference, however, is that the center of oscillation does not correspond to the position of neutral spring configuration; rather it is shifted down by a vertical length given by :

$⇒{y}_{0}=\frac{mg}{k}$

## Block connected to springs in series

We consider two springs of different spring constants. An external force like gravity produces elongation in both springs simultaneously. Since spring is mass-less, spring force is same everywhere in two springs. This force, however, produces different elongations in two springs as stiffness of springs are different. Let “ ${y}_{1}$ ” and “ ${y}_{2}$ ” be the elongations in two springs. As discussed for the single spring, the net restoring force for each of the springs is given as :

${F}_{\text{net}}=-{k}_{1}{y}_{1}=-{k}_{2}{y}_{2}$ The spring is stretched a bit from the equilibrium position and then let go to oscillate.

The total displacement of the block from equilibrium position is :

$⇒y={y}_{1}+{y}_{2}=-\frac{{F}_{\text{net}}}{{k}_{1}}-\frac{{F}_{\text{net}}}{{k}_{2}}$

$⇒{F}_{\text{net}}=\frac{{k}_{1}{k}_{2}y}{{k}_{1}+{k}_{2}}$

A comparison with the expression of extension of the single spring at equilibrium position reveals that spring constant of the arrangement of two springs is equivalent to a single spring whose spring constant is given by :

$⇒k=\frac{{k}_{1}{k}_{2}}{{k}_{1}+{k}_{2}}$

This relationship can also be expressed as :

$⇒\frac{1}{k}=\frac{1}{{k}_{1}}+\frac{1}{{k}_{2}}$

In the nutshell, we can consider the arrangement of two springs in series as a single spring of spring constant “k”, which is related to individual spring constants by above relation. Further, we can extend this concept to a number of springs by simply extending the relation as :

$⇒\frac{1}{k}=\frac{1}{{k}_{1}}+\frac{1}{{k}_{2}}+\frac{1}{{k}_{3}}+\dots$

The periodic attributes are given by the same expressions, which are valid for oscillation of single spring. We only need to use equivalent spring constant in the expression.

## Block in between two springs

In this arrangement, block is tied in between two springs as shown in the figure. In order to analyze oscillation, we consider oscillation from the reference position of equilibrium. Let the block is displaced slightly in downward direction (reasoning is similar if block is displaced upward). The upper spring is stretched, whereas the lower spring is compressed. The spring forces due to either of the springs act in the upward direction. The net downward displacement is related to net restoring force as :

${F}_{\text{net}}=-{k}_{1}{y}_{1}-{k}_{2}{y}_{2}=-\left({k}_{1}+{k}_{2}\right)y$ The spring is stretched a bit from the equilibrium position and then let go to oscillate.

A comparison with the expression of extension of the single spring reveals that spring constant of the arrangement of two springs is equivalent to a single spring whose spring constant is given by :

$⇒k={k}_{1}+{k}_{2}$

Clearly, the periodic attributes are given by the same expressions, which are valid for oscillation of single spring. We only need to use equivalent spring constant in the expression.

## Block connected to springs in parallel

Here, we consider a block is suspended horizontally with the help of two parallel springs of different spring constants as shown in the figure. When the block is pulled slightly, it oscillates about the equilibrium position. The net restoring force on the block is :

${F}_{\text{net}}=-{k}_{1}{y}_{1}-{k}_{2}{y}_{2}=-\left({k}_{1}+{k}_{2}\right)y$ The spring is stretched a bit from the equilibrium position and then let go to oscillate.

A comparison with the expression of extension of the single spring reveals that spring constant of the arrangement of two springs is equivalent to a single spring whose spring constant is given by :

$⇒k={k}_{1}+{k}_{2}$

Again, the periodic attributes are given by the same expressions, which are valid for oscillation of single spring. We only need to use equivalent spring constant in the expression.

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