Question Video: Relating the Frequency of an Oscillating Spring to Its Load Mass

A mass π‘šβ‚€ is attached to a spring and hung vertically. The mass is raised a short distance in the vertical direction and released. The mass oscillates with a frequency 𝑓₀. If the mass is replaced with a mass nine times as large, and the experiment was repeated, what would be the frequency of the oscillations in terms of 𝑓₀?

02:47

Video Transcript

A mass π‘š zero is attached to a spring and hung vertically. The mass is raised a short distance in the vertical direction and released. The mass oscillates with a frequency 𝑓 sub zero. If the mass is replaced with a mass nine times as large and the experiment was repeated, what would be the frequency of the oscillations in terms of 𝑓 sub zero?

In this problem, we want to solve for the frequency of the oscillations of the mass with mass nine times π‘š sub zero. We’ll call that frequency 𝑓. Let’s draw a diagram that shows the experiment being described.

We begin with a spring hanging vertically from a ceiling. A mass of value π‘š sub zero is attached to the end of the spring. And when this mass is released, the mass oscillates on the end of the spring up and down at a frequency of 𝑓 sub zero.

Then we put a different mass on the end of the spring, this one with nine times the value of the original mass. When we release hold of this mass on the end of the spring, the mass again oscillates at some new frequency we’ve called 𝑓. We want to solve for 𝑓 in terms of 𝑓 sub zero, the original frequency of oscillation.

Let’s recall the mathematical relationship for the frequency of an oscillating spring. The frequency of a spring’s oscillation is equal to one over two πœ‹ times the square root of the spring constant π‘˜ divided by the mass on the end of the spring.

In the case of our original mass, π‘š sub zero, and our original frequency, 𝑓 sub zero, we can write that 𝑓 sub zero is equal to one over two πœ‹ times the square root of π‘˜, the unknown spring constant, divided by π‘š sub zero.

Now let’s write a new spring frequency equation using 𝑓, the frequency we want to solve for, and nine times π‘š zero in place of π‘š zero.

Looking at this second equation for spring oscillation frequency, we see that we can factor out a one-third term from underneath the square root and that when we do that the resulting terms left inside parentheses are equal to 𝑓 sub zero, the original oscillation frequency of the spring.

Therefore, 𝑓, the springs oscillation frequency when the mass has been multiplied by nine, is equal to one-third the oscillation frequency when the mass was equal to π‘š sub zero.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.