Simple Harmonic Motion of a Mass on a Spring

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Homework Help Overview

The discussion revolves around a mass-spring system undergoing simple harmonic motion. The original poster presents a series of calculations related to the system's force constant, frequency, maximum speed, acceleration, total energy, and specific conditions of speed and acceleration at one-third of the maximum displacement.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore various calculations related to simple harmonic motion, including the force constant and energy. Some question the correctness of specific calculations, particularly for letters f and g, while others discuss potential alternative approaches for determining acceleration in relation to distance.

Discussion Status

Some participants have provided feedback on the calculations, indicating that the steps are generally acceptable. There is an ongoing exploration of whether simpler methods exist for certain calculations, particularly for letter g, but no consensus has been reached on the best approach.

Contextual Notes

Participants express uncertainty about the existence of equations that relate acceleration directly to distance, which adds complexity to the discussion of letter g.

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A 2 kg mass is attached to a spring aand placed on a horizontal, smooth surface. A force of 20N is required to hold the mass at reast when it is pulled .200m from the equilibrium position. The mass is released and undergoes simple harmoic motion.
a) what is the force constant?
b) frequency?
c) max speed and where does this occur?
d) mass acceleration and where does this occur?
e) total energy
f) speed when displacement equals one third of the max value
g) acceleration when displacement equals one third of the max value

a)
N/m=20/0.2m=100N/m

b)
[tex]T=2\pi\sqrt{\frac{m}{k}}[/tex]
[tex]T=2\pi\sqrt{\frac{2kg}{100N/m}}[/tex]
[tex]T=0.8886s[/tex]

[tex]f=1/T=1.1125Hz[/tex]

c)
[tex]v_{max}=\sqrt{\frac{m}{k}}*A[/tex]
[tex]v_{max}=\sqrt{\frac{2kg}{100N/m}}*0.2m[/tex]
[tex]v_{max}=0.0283m/s@x=0m[/tex]

d)
[tex]a_{max}=\frac{mA}{k}}[/tex]

[tex]a_{max}=\frac{2kg*0.2m}{100N/m}}[/tex]
[tex]a_{max}=0.004m/s^2@x=+/-0.2m[/tex]

e)
[tex]E=0.5kA^2[/tex]
[tex]E=0.5(100N/m)(0.2m)^2[/tex]
[tex]E=2J[/tex]

f)
[tex]d_{max}/3=0.2/3=\frac{1}{15}m[/tex]
[tex]v=\sqrt{\frac{k}{m} (A^2-x^2)}[/tex]
[tex]v=\sqrt{\frac{100N/m}{2kg} ((0.2m)^2-(\frac{1}{15}m)^2)}[/tex]
[tex]v=+/- \frac{4}{3}m/s[/tex]

g)
[tex]v=-\omega A sin \omega t[/tex]

[tex]\frac{4}{3}m/s=-\sqrt{\frac{100N/m}{2kg}} *0.2m sin \sqrt{\frac{100N/m}{2kg}} t[/tex]

[tex]\frac{4}{3}m/s=-1.414213562sin7.071067812t[/tex]

[tex]t=-0.1740839504s[/tex]

[tex]a=-\omega^2 A cos \omega t[/tex]

[tex]a=- \frac{100N/m*0.2m}{2kg} cos 7.071067812*-0.1740839504s[/tex]

[tex]a=+/- \frac{10}{3} m/s^2[/tex]

did I do this right? especially letter (g) thanks
 
Last edited:
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could someone please check just letter f and g for me? thanks
 
The steps are fine. I didn't checked the calculations.
 
I just didnt know if there was an easier way to do letter g. For letter f, there is an equation that has both velocity and distance, but for g, I don't know if there is an equation that has acceleration in terms of distance.

this is why I'm hesitant on my way of answering letter g
 
any help??
 
I don't know of such an equation, but it doesn't matter. As long as you have A way to do it.
 
You used the correct formula on f. For g) a simpler method would be to use F=kx, so a=kx/m. Honestly can't be troubled to check your method.
 

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