Vertical Spring - Block Oscillator

A mass attached at the end of a vertical spring executes simple harmonic motion. When an object is attached at the end of a spring which is in equilibrium position, the object exerts a downward force pulling the spring down and a restoring force from the spring pulls the object back to equilibrium position. This up and down motion repeats in a period of time and this is when the simple harmonic motion happens.

In equilibrium position, the net force acting on the object is zero. That is, as shown by the free-body diagram, $$F_g=mg$$ and from Hooke's Law, the force exerted by the spring is $$F_s=-ky$$, $$y$$ is used indicating that the object is in vertical motion. Thus,

$$F_{ynet }=F_g+F_s=mg-ky=0 \implies mg=ky$$.

As the object applied force on the spring, the spring is stretched downward to maximum displacement, $$+A$$. The displacement of the spring will become $$y+A$$. Thus, the total force can be expressed as

$$F_{ynet }=F_g+F_s=mg-k(y+A)=mg-ky-kA$$.

Since $$mg-ky=0$$, this implies that the net force is

$$F_{ynet}=-kA$$.

This is also true when the spring pulls the mass upward.

In a spring-block oscillator, the mass of the object and the type of spring affects the simple harmonic motion. An increase in the mass indicates an increase in the period and a stiffer spring indicates decrease in period. Thus, the period in a vertical spring-block oscillator can be calculated as

$$T=2\pi \sqrt{\frac {m}{k}}$$

Example 1.

A spring in equilibrium position is stretched due to an object with a mass of $$3.0\;kg$$ attached to it. If the spring constant is $$300\;N/m$$, what is the period of the oscillating object?

Given:

$$m=3.0\;kg\\ k=300\;N/m$$

Solution:

Calculate the period using the equation $$T=2\pi \sqrt{\frac {m}{k}}$$.

$$T=2\pi \sqrt{\frac {3.0\;kg}{300\;N/m}}=2\pi(0.1\;s)=0.63\;s$$

Example 2.

Andres attached his toy car to a slinky toy hanged on a brach of a tree. If the slinky displaced to $$1 \;m$$ after the toy car was attached to it, what is the mass of the car? How long does it take to complete one cycle of the up and down motion of the car? The spring constant is $$0.98\;N/m$$.

Given:

$$A=1.0\;m\\ k=0.98\;N/m$$

Solution 1:

First step is to solve for the spring force using the Hooke's Law $$F=ky$$, where $$y=A=1.0\;m$$.

$$F_s=(0.98\;N/m)(1.0\;m)=0.98\;N$$.

Solution 2:

Since $$F_s=F_g=0.98\;N$$, we can solve for the mass of the object using $$F_g=mg$$.

$$m=\frac{F_g}{g}=\frac{0.98\;N}{9.8\;m/s^2}=0.1\;kg$$.

Hence, the mass of the object is $$0.1\;kg$$.

Solution 3.

Period can now be solved using the equation $$T=2\pi \sqrt{\frac {m}{k}}$$.

$$T=2\pi \sqrt{\frac {0.1\;kg}{0.98\;N/m}}=2\pi(0.32\;s)=2.01\;s$$.

Therefore, it takes $$2.01\;s$$ to complete one cycle.