Spring-Block Oscillator: Vertical Motion, Frequency & Mass

An error occurred trying to load this video.

Try refreshing the page, or contact customer support.

Coming up next: Pendulums in Physics: Definition & Equations

You're on a roll. Keep up the good work!

Take Quiz Watch Next Lesson
Your next lesson will play in 10 seconds
  • 0:01 What is a Spring-Block…
  • 1:06 Equations
  • 2:27 Example Problem
  • 3:25 Lesson Summary
Add to Add to Add to

Want to watch this again later?

Log in or sign up to add this lesson to a Custom Course.

Login or Sign up

Create an account to start this course today
Try it free for 5 days!
Create An Account

Recommended Lessons and Courses for You

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.


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.

To unlock this lesson you must be a Study.com Member.
Create your account

Register for a free trial

Are you a student or a teacher?
I am a teacher
What is your educational goal?

Unlock Your Education

See for yourself why 10 million people use Study.com

Become a Study.com member and start learning now.
Become a Member  Back

Earning College Credit

Did you know… We have over 95 college courses that prepare you to earn credit by exam that is accepted by over 2,000 colleges and universities. You can test out of the first two years of college and save thousands off your degree. Anyone can earn credit-by-exam regardless of age or education level.

To learn more, visit our Earning Credit Page

Transferring credit to the school of your choice

Not sure what college you want to attend yet? Study.com has thousands of articles about every imaginable degree, area of study and career path that can help you find the school that's right for you.

Create an account to start this course today
Try it free for 5 days!
Create An Account