How Do You Calculate Mass Using Spring Constant and Frequency?

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To calculate mass using the spring constant and frequency, the formula m = k/(2πf)² is derived, where k is the spring constant and f is the frequency in Hz. For a spring with a constant of 4 N/m and a frequency of 0.8 Hz, the mass is calculated to be approximately 0.158 kg. When friction is introduced and the spring constant is unknown, two reference weights yield different spring constant values, indicating a potential issue with measurements or calculations. The discussion highlights the importance of using consistent units and considering damping effects in calculations. Accurate determination of mass requires careful calibration and consideration of all variables involved.
SithsNGiggles
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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!
 
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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:
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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