Simple harmonic motion and oscillation

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Homework Help Overview

The discussion revolves around a problem involving simple harmonic motion of a mass attached to a spring. The mass is set into motion after being lowered to its equilibrium position, and participants are exploring various aspects such as frequency, period, maximum velocity, and maximum acceleration of the oscillating mass.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the equations related to simple harmonic motion, including the relationship between mass, spring constant, and angular frequency. There are questions about the meaning of the spring constant (k) and its relation to forces acting on the mass. Some participants express uncertainty about how to determine the amplitude and maximum velocity without initial conditions.

Discussion Status

The discussion is active with participants providing equations and clarifications regarding the concepts involved. Some have made progress in calculating the spring constant and period, while others are still grappling with the implications of missing initial conditions for determining amplitude and maximum velocity.

Contextual Notes

Participants note the lack of initial conditions, such as the initial position and velocity of the mass when set into motion, which affects the ability to find numerical answers for certain quantities like amplitude.

warmfire540
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A 780-g mass is attached to a vertical spring and lowered slowly until the mass rests at its equilibrium position 30 cm below the original length of the spring. It is then set into simple harmonic motion.

(a) What is the frequency and period of the oscillation?
(b) What is the maximum velocity of the oscillating mass? Where does the maximum velocity occur?
(c) What is the maximum acceleration of the oscillating mass? Where does the maximum acceleration occur?
(d) Draw a diagram showing how the total energy of the oscillating system changes with time.

Can someone get me off to a start here? I don't understand how to find the frequency without knowing Lamda or angular frequence (w)..
 
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What are the equations that you know? We have to know what you know, so that we can pick out what you are struggling with. Maybe to get you started

[tex]m\ddot{x} + kx=0[/tex]
leads to
[tex]x(t)=A_0 cos(\omega t + \delta)[/tex]

where [itex]\omega = \sqrt{k/m}[/itex]. So now you know the angular frequency.
 
Mindscrape said:
What are the equations that you know? We have to know what you know, so that we can pick out what you are struggling with. Maybe to get you started

[tex]m\ddot{x} + kx=0[/tex]
leads to
[tex]x(t)=A_0 cos(\omega t + \delta)[/tex]

where [itex]\omega = \sqrt{k/m}[/itex]. So now you know the angular frequency.



Yeah, but what is k? The spring constant? or does it have another meaning?
 
I just don't get the k part..
 
k is the spring constant make a free body diagram and you'll see!
 
Large value of k means that the spring resists being stretched, a very hard spring. A small value of k means the spring is very soft and can be stretched quite easily. Restoring force of a spring is F= -k(delta s). Opposite in direction to the displacement.
 
habibclan said:
Large value of k means that the spring resists being stretched, a very hard spring. A small value of k means the spring is very soft and can be stretched quite easily. Restoring force of a spring is F= -k(delta s). Opposite in direction to the displacement.


I still don't see.. I only know the weight, and the distance from it's equillibrium to rest position.
 
Tell me, if you had to associate familiar physics term with weight, what would it be?
 
Mindscrape said:
Tell me, if you had to associate familiar physics term with weight, what would it be?

Huh? W=ma?
 
  • #10
Weight is a force. Are spring constants associated with a force at all? What does it mean when something is in equilibrium?
 
  • #11
Mindscrape said:
Weight is a force. Are spring constants associated with a force at all? What does it mean when something is in equilibrium?



Oh, i see, well the weight acting down is .78*9.8=7.644 (mg)
that means there must be an equal force acting up which must be 7.644 in the opposite direction. So is this equal force the spring constant? (k)
 
  • #12
A force constant like k expresses the force generated per unit of some variable (whether it be displacement or acceleration)
 
  • #13
-kx=mg

in other words
 
  • #14
okay, so k(-.3)=7.644
k=25.48

T=2pi*sqrt(m/k) (period of a loaded spring)
T=2pi*sqrt(.78/25.48)
T=1.099

f=1/T
f=.9096
 
  • #15
warmfire540 said:
okay, so k(-.3)=7.644
k=25.48

T=2pi*sqrt(m/k) (period of a loaded spring)
T=2pi*sqrt(.78/25.48)
T=1.099

f=1/T
f=.9096

just like that?
 
  • #16
Just like that.
 
  • #17
Mindscrape said:
Just like that.

Awesome!

for part b, the max velocity is:

well, i know two equations

vmax=sqrt(k/m)A
vmax=wA (w being angular velocity)
however, i also know that velocity is greatest at the equilibrium position
so how do i find this? do i need to know k?

time out, i know k..k=25.48
so vmax=sqrt(25.48/.78)A
vmax=5.715(A) but is A (the amplitude) the 30cm?
 
  • #18
You won't be able to find a numerical answer for the amplitude with the information provided. You would need to know it's initial position and initial velocity when it was first set into harmonic motion.

An algebraic answer should be fine.
 
  • #19
Mindscrape said:
You won't be able to find a numerical answer for the amplitude with the information provided. You would need to know it's initial position and initial velocity when it was first set into harmonic motion.

An algebraic answer should be fine.

so just leave it at:
vmax=5.715(A)
 
  • #20
Yep, I don't see anyway around the lack of initial conditions.
 

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