Simple Harmonic motion of vertical spring.

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Homework Help Overview

The problem involves an object of unknown mass suspended from a vertical spring with an unknown spring constant. The object is at rest when the spring extends by 14 cm and then begins to oscillate in simple harmonic motion (SHM) after being pushed. The goal is to determine the period of the oscillation.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the equilibrium condition where the forces of gravity and spring force balance. There is an attempt to derive the angular frequency and period from the equilibrium position. Some participants question the calculations related to the angular frequency and the square root involved.

Discussion Status

Participants are actively engaging with the calculations, with one participant acknowledging a mistake in their computation of the angular frequency. There is a recognition of the need to check for basic calculation errors before delving into more complex physics concepts.

Contextual Notes

There is an indication that the original poster is struggling with similar problems, suggesting a pattern of misunderstanding that may need to be addressed. The discussion reflects a learning process where foundational errors are being identified.

Kizaru
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Homework Statement


An object of unknown mass M is hung from a vertical spring of unknown spring constant k, and the object is observed to be at rest when the spring has extended by 14cm. The object is then given a slight push and executes SHM. Determine the period of this oscillation.


Homework Equations


T = 1/f
w = (k/M)^(1/2) = 2pi*f
F=kx

The Attempt at a Solution


The object is hanging, so it has a force of Mg pulling it down and kx pulling it up. It's in equilibrium at a displacement of 14cm (0.14m). So I set kx = Mg and solved for k:
k = Mg / x

I then plugged this into w = (k/M)^(1/2) so the M's cancel and I'm left with w = (g/x)^(1/2) = (9.8 / 0.14)^(1/2) = 70rad/s

w = 2pi*f => f = w/(2pi) = 11.14 Hz
T = 1/f = 0.089s

This isn't the right answer according to the book, and I keep getting this kind of problem wrong. I'm not entirely sure what I'm doing wrong.
 
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The Attempt at a Solution




I then plugged this into w = (k/M)^(1/2) so the M's cancel and I'm left with w = (g/x)^(1/2) = (9.8 / 0.14)^(1/2) = 70rad/s


Check the above calculation
 
Hi Kizaru! :smile:

(have a square-root: √ and an omega: ω :wink:)
Kizaru said:
w = (g/x)^(1/2) = (9.8 / 0.14)^(1/2) = 70rad/s

Nooo … √70 :wink:
 
Ah yes, turns out my mistake was ignoring the square root. Thanks. I made that same mistake for all of these types of problems hehe, thanks :) Embarassing, guess I should check for elementary mistakes before I look for physics mistakes :)
 

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