Simple Harmonic Motion of a Mass on a Spring

In summary, a 2 kg mass is attached to a spring and placed on a horizontal, smooth surface. A force of 20N is required to hold the mass at rest when it is pulled 200m from the equilibrium position. The mass is released and undergoes simple harmonic motion.
  • #1
UrbanXrisis
1,196
1
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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  • #2
could someone please check just letter f and g for me? thanks
 
  • #3
The steps are fine. I didn't checked the calculations.
 
  • #4
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
 
  • #5
any help??
 
  • #6
I don't know of such an equation, but it doesn't matter. As long as you have A way to do it.
 
  • #7
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.
 

1. What is Simple Harmonic Motion?

Simple Harmonic Motion (SHM) is a type of periodic motion in which the restoring force is directly proportional to the displacement from equilibrium and acts in the opposite direction to the displacement. This results in a back-and-forth motion around the equilibrium point.

2. What is a Mass on a Spring?

A mass on a spring is a physical system in which a mass is attached to a spring and is free to move back and forth. The spring provides the restoring force for the mass, causing it to undergo SHM.

3. What factors affect the frequency of a mass on a spring?

The frequency of a mass on a spring is affected by the mass of the object, the stiffness of the spring, and the gravitational acceleration. In general, a heavier mass, stiffer spring, and larger gravitational acceleration will result in a higher frequency.

4. What is the equation for the period of a mass on a spring?

The period (T) of a mass on a spring can be calculated using the equation T = 2π * √(m/k), where m is the mass of the object and k is the spring constant (a measure of the stiffness of the spring).

5. How does the amplitude affect the motion of a mass on a spring?

The amplitude of a mass on a spring is the maximum displacement from equilibrium that the mass undergoes. The amplitude affects the maximum speed and acceleration of the object, but does not affect the frequency or period of the motion.

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