# How Do You Calculate Mass Using Spring Constant and Frequency?

• SithsNGiggles
In summary: I found, in blue.In summary, the conversation discusses using a spring with a spring constant of 4 N/m to weigh items, assuming no friction. In part (a), a frequency of 0.8 Hz is given and the mass is calculated to be 0.158 kg. In part (b), a formula for the mass in terms of frequency is derived. In part (2), friction is added and the spring constant is unknown. Two reference weights of 1 kg and 2 kg are used to calibrate the setup, with measured frequencies of 1.1 Hz and 0.8 Hz, respectively. In part (a), the values for the spring constant k and damping constant c are found
SithsNGiggles

## Homework Statement

(1) Suppose you have a spring with spring constant 4 N/m. You want to use it to weigh items. Assume no friction. You place the mass on the spring and put it in motion.
a) You count and find that the frequency is 0.8 Hz (cycles per second). What is the mass?
b) Find a formula for the mass m given the frequency ω in Hz.​

(2) Suppose we add possible friction to the previous situation. Further, suppose the spring constant is unknown, but you have two reference weights 1 kg and 2 kg to calibrate your setup. You put each in motion on your spring and measure the frequency. For the 1 kg weight you measured 1.1 Hz and for the 2 kg weight you measured 0.8 Hz.
a) Find k (spring constant) and c (damping constant).
b) Find a formula for the mass in terms of the frequency in Hz.
c) For an unknown object you measured 0.2 Hz, what is the mass of the object? Suppose that you know the mass of the unknown object is more than a kilogram.​

## Homework Equations

##\omega=\sqrt{\frac{k}{m}}##

## The Attempt at a Solution

• I'm not sure if this is right, but for 1(a) I use the above formula and I get
##0.8=\sqrt{\frac{4}{m}}\\
m=6.25##
• For 1(b), here's what I did:
##\omega=\sqrt{\frac{k}{m}}\Rightarrow m=\frac{k}{\omega^2}##
But this seems too simple.
• For 2(a), I tried to find k using the formula under "Relevant equations," but I get two different values of k:
##1.1=\sqrt{\frac{k}{1}}\Rightarrow k=1.21\\
0.8=\sqrt{\frac{k}{2}}\Rightarrow k=1.28##
So I'm not really sure what's going on here.

Any input is appreciated, thanks!

SithsNGiggles said:

## Homework Statement

(1) Suppose you have a spring with spring constant 4 N/m. You want to use it to weigh items. Assume no friction. You place the mass on the spring and put it in motion.
a) You count and find that the frequency is 0.8 Hz (cycles per second). What is the mass?
b) Find a formula for the mass m given the frequency ω in Hz.​

(2) Suppose we add possible friction to the previous situation. Further, suppose the spring constant is unknown, but you have two reference weights 1 kg and 2 kg to calibrate your setup. You put each in motion on your spring and measure the frequency. For the 1 kg weight you measured 1.1 Hz and for the 2 kg weight you measured 0.8 Hz.
a) Find k (spring constant) and c (damping constant).
b) Find a formula for the mass in terms of the frequency in Hz.
c) For an unknown object you measured 0.2 Hz, what is the mass of the object? Suppose that you know the mass of the unknown object is more than a kilogram.​

## Homework Equations

##\omega=\sqrt{\frac{k}{m}}##

## The Attempt at a Solution

• I'm not sure if this is right, but for 1(a) I use the above formula and I get
##0.8=\sqrt{\frac{4}{m}}\\
m=6.25##
• For 1(b), here's what I did:
##\omega=\sqrt{\frac{k}{m}}\Rightarrow m=\frac{k}{\omega^2}##
But this seems too simple.
• For 2(a), I tried to find k using the formula under "Relevant equations," but I get two different values of k:
##1.1=\sqrt{\frac{k}{1}}\Rightarrow k=1.21\\
0.8=\sqrt{\frac{k}{2}}\Rightarrow k=1.28##
So I'm not really sure what's going on here.

Any input is appreciated, thanks!

Not right. For one the units of ω in your formula aren't Hz. They are radians/s. There is some difference between them. For the second part there is another formula for ω that include the effect of damping. Try to look it up!

SithsNGiggles said:
• I'm not sure if this is right, but for 1(a) I use the above formula and I get
##\color{red}{5.027}=\sqrt{\frac{4}{m}}##
##\color{red}{m=0.158}##
• For 1(b), here's what I did:
##x=\text{ frequency given in Hz}##
##2\pi x=\text{ given frequency in rad/s}##
##x=\frac{1}{2\pi}\sqrt{\frac{k}{m}}\Rightarrow m=\frac{k}{(2\pi x)^2}##
• For 2(a), I tried to find k using the formula under "Relevant equations," but I get two different values of k:
##1.1=\sqrt{\frac{k}{1}}\Rightarrow k=1.21\\
0.8=\sqrt{\frac{k}{2}}\Rightarrow k=1.28##
So I'm not really sure what's going on here.
Ah, thanks, I've corrected the first part, in red. There doesn't appear to be any other formula for ω in my notes, though. I'll be on the lookout for it.

EDIT: also corrected 1(b)

Last edited:

## What is a spring system differential equation?

A spring system differential equation is a mathematical representation of the motion of a mass attached to a spring. It takes into account the properties of the spring, such as its stiffness, and the external forces acting on the mass.

## What is the equation for a spring system?

The equation for a spring system is typically written as m*d^2x/dt^2 + kx = F, where m is the mass of the object, x is its displacement from equilibrium, k is the spring constant, and F is the external force acting on the object.

## How do you solve a spring system differential equation?

To solve a spring system differential equation, you can use techniques from differential equations, such as separation of variables or the method of undetermined coefficients. You can also use the equation of motion approach to find the general solution.

## What factors influence the behavior of a spring system?

The behavior of a spring system is influenced by several factors, including the mass and stiffness of the spring, the initial conditions of the system, and any external forces acting on the system. Temperature and damping can also affect the behavior of a spring system.

## What are some real-life applications of spring system differential equations?

Spring system differential equations have many real-life applications, such as in modeling the motion of objects attached to springs, analyzing the vibrations of mechanical systems, and understanding the behavior of biological systems like the human cardiovascular system.

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