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.