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


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



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


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


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?


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

Solution 1:

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


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.


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

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.