# Spring-Block Oscillator: Vertical Motion, Frequency & Mass

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• 0:01 What is a Spring-Block…
• 1:06 Equations
• 2:27 Example Problem
• 3:25 Lesson Summary

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Lesson Transcript
Instructor: David Wood

David has taught Honors Physics, AP Physics, IB Physics and general science courses. He has a Masters in Education, and a Bachelors in Physics.

After watching this video, you will be able to explain how a spring-block oscillator is an example of simple harmonic motion. You will also be able to use an equation for a spring-block oscillator that relates the frequency and mass of the block.

## What Is a Spring-Block Oscillator?

A spring-block oscillator is where you hang a block of mass m on a vertically hanging spring, stretch it, and then let it bounce back and forth. This bouncing is an example of simple harmonic motion.

Simple harmonic motion is any motion where a restoring force is applied that is proportional to the displacement, in the opposite direction of that displacement. Or in other words, the more you pull it, the more it wants to go the opposite way, back towards the center. The spring-block oscillator is an example of this, because the more you stretch it, the more force you feel back towards the equilibrium position.

A spring-block oscillator involves the interplay between three types of energy: gravitational potential energy at the top, elastic potential energy at the bottom, and kinetic energy in the middle. This makes it a somewhat more complicated example of simple harmonic motion, compared to a horizontal spring or a pendulum.

In this lesson, we're going to analyze this particular example of simple harmonic motion and look at equations involving the frequency of the oscillation and the mass of the block.

## Equations

Simple harmonic motion has a variety of equations. But for a mass on a spring, we can use this equation to calculate the time period:

T is the time period of the oscillation, measured in seconds, and this is equal to 2pi times the square-root of m over k, where m is the mass of the object connected to the spring measured in kilograms, and k is the spring constant of the spring. The spring constant is a number that represents how stretchy the spring is. A bigger spring constant means the spring is stiffer.

Frequency and time period are the inverse of each other, which means that time period is equal to one over the frequency. We can therefore adjust this equation to include the frequency, measured in Hertz. The frequency tells us the number of oscillations per second, so a frequency of 120 Hz means that the full motion repeats 120 times every second.

This equation shows us that if you increase the mass on the spring, you increase the time period, thereby decreasing the frequency of the oscillation. And if you increase the spring constant (or the stiffness) of the spring, then you decrease the time period, which in turn increases the frequency of the oscillation. Because of the square-root sign, this isn't a direct proportion relationship. Doubling m or k will cause the time period and frequency to change by less than a factor of 2.

## Example Problem

Okay, let's go through an example. Let's say you have a mass of 6 kilograms attached to a spring, with spring constant 4 N/m. What is the time period and frequency of the oscillation?

First of all, as always, we should write down what we know. The mass on the spring, m, equals 6 kg. The spring constant, k, is 4 N/m. And we're trying to find f and T.

All we have to do is plug our numbers into the equation and solve. We can solve for T or solve for f, because they're just the reciprocal of each other. Once we have one, we just have to calculate 1 over the other, and we'll have both our numbers.

So let's use the straightforward time period equation. Plugging in and solving gives us a time period of 7.7 seconds. To find our frequency, calculate 1 divided by 7.7, and that gives us 0.13 Hz. And that's it; we're done.

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