**Spring Combinations - Block Oscillator**

Simple harmonic motion can also be demonstrated by a block attached to two springs which are in parallel or in series with each other.

Two springs connected in parallel to a block with mass, \(m\). The displacement of the block is said to be equal if the springs are parallel to each other. Thus, given two springs of different constants, \(k_1\) and \(k_2\), and the block displaces to \(x\), the spring force is expressed as

\(F=k_1x+k_2x=(k_1+k_2)x\)

Denote \(k_1+k_2\) as \(k_{eq}\) which is the total spring constant, the above equation will become

\(F=k_{eq}x\).

When two springs are connected in series and a block with mass, \(m\) is attached at the end, the force exerted by the springs on the block is the same. Mathematically, it can be expressed as

\(F=-k_1x_1=-k_2x_2\)

The total spring constant, \(k_{eq}\) of series springs is

\(\frac{1}{k_{eq}}=\frac{1}{k_1}+\frac{1}{k_2}\).

**Example 1.**

A \(30-kg\) mass is attached to two springs in parallel. If spring 1 has a constant of \(50\; N/m\) and spring constant 2 is \(40\;N/m\), what is the period of oscillation of the block?

Given:

\(m=30\;kg\\ k_1=50\;N/m\\ k_2=40\;N/m\)

Solution 1:

Solve first for the equivalent spring constant using the equation \(k_{eq}=k_1+k_2\).

\(k_{eq}=(50\;N/m)+(40\;N/m)=90\;N/m\)

Solution 2:

Solve for the period of oscillation using the formula \(T=2\pi \sqrt{\frac{m}{k}}\).

\(T=2\pi \sqrt{\frac{30\;kg}{90\;N/m}}=2\pi(0.58)=3.64\;s\).

Thus, the object oscillates completely at \(3.64\;s\).

**Example 2.**

Calculate for the period of oscillation of the block connected to two springs as shown in the figure.

Given:

\(k_1=12\;N/m\\ k_2=6\;N/m\)

Solution 1: Solve for the total spring constants of springs in series by the equation \(\frac{1}{k_{eq}}=\frac{1}{k_1}+\frac{1}{k_2}\).

\( \frac{1}{k_{eq}}=\frac {1}{12}+\frac{1}{6}=\frac{3}{12}=\frac{1}{4}\)

Thus, \(k_{eq}=4\;N/m\).

Solution 2:

The period of oscillation is solved using the equation for period \(T=2\pi \sqrt {\frac{m}{k}}\).

**\(T=2\pi \sqrt {\frac{5\;kg}{4\;N/m}}=2\pi(1.12)=7.04\;s\).**

Therefore, it takes \(7.04\;s\) to complete one cycle in simple harmonic motion.