# Simple Harmonic Motion-Solving for Spring Constant

1. Jan 14, 2010

### Wellesley

1. The problem statement, all variables and given/known data
16. Suppose that a particle of mass 0.24 kg acted upon by a spring undergoes simple harmonic motion with the parameters given in Problem 1.

(a) What is the total energy of this motion?

(b) At what time is the kinetic energy zero? At what time is the potential energy zero?

(c) At what time is the kinetic energy equal to the potential energy?
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1. A particle moves back and forth along the x axis between the points x = 0.20 m and x = –0.20 m. The period of the motion is 1.2 s, and it is simple harmonic. At the time t = 0, the particle is at x = 0 and its velocity is positive.

(a) What is the frequency of the motion? The angular frequency?

(b) What is the amplitude of the motion?

(c) What is the phase constant?

(d) At what time will the particle reach the point x = 0.20 m? At what time will it reach the point x = –0.10 m?

(e) What is the speed of the particle when it is at x = 0? What is the speed of the particle when it reaches the point x = – 0.10 m?

2. Relevant equations
f=frequency=0.83 Hz
$$\omega$$= angular frequency= 2*pi*0.3= (5*pi)/3 or 5.23599 rad/s
T=period=1.2 s
m=mass=0.24 kg
A= amplitude=0.2 m

3. The attempt at a solution

My problem lies in part a of number 16: What is the total energy of this motion? The rest of the question I can solve easily.
I know the error has to do with the spring constant and the angular frequency.

$$\omega=$$$$\sqrt{k/m}$$
k=$$\omega^{2}$$ * m
k=5.2359992*0.24=6.57974 N/m ----> [STRIKE]Pretty darn small for a spring constant.[/STRIKE]
I then tried this to make sure k was right:
T=2*pi*$$\sqrt{k/m}$$
And got: k=6.57974 N/m

Then when I solve for total energy....
E=1/2kA2 -->1/2*6.57974 N/m *(0.2m)2--->E=.131595 J
The answer in the back is 6.6 Joules.

When I tried to work backwards, from the answer I get:

(6.6 J*2)/.22-->330 N/m as the spring constant.

I'm pretty sure my numbers are right....what could be throwing the answer off by a factor of fifty times? From past experience, I'm pretty certain the book isn't wrong, but in this case, I'm not sure............
Any help would be appreciated. Thanks!

Last edited: Jan 15, 2010
2. Jan 15, 2010

### CFDFEAGURU

The units produced from your equation for k produce kg/s^2. That is not a spring constant unit. The spring constant is a force/distance.

Also, you only calculated the potential energy, the total energy of the system involves the kinetic energy as well.

E = T + V

where E is the total energy
T is the kinetic energy
V is the potential energy

First calculate the spring constant with the correct units.

Thanks
Matt

3. Jan 15, 2010

### Wellesley

I know that the units for my spring constant are correct. Newton/meter--> (kg*m/s^2)/meter ---->leaves kg/s^2 as the units. If you were to multiply kg/s2 by a meter, you would get kg*m/s2, which is a newton (and the correct unit from the equation F=-kx).
This link also confirms this....http://www.rwc.uc.edu/koehler/biophys.2ed/sho.html [Broken]
I'm confused. I did calculate the total energy. I used this equation: E=1/2kA2...the same one from Wikipedia (and my physics book).......

http://en.wikipedia.org/wiki/Simple_harmonic_motion ---> (under Energy of simple harmonic motion).

After going through the equations again, I still feel my answer is right. Any thoughts anybody?

Last edited by a moderator: May 4, 2017
4. Jan 15, 2010

### ehild

ehild

5. Jan 15, 2010

### Wellesley

Thanks. That makes me feel a little better. While I was doing this problem yesterday, I thought I was losing it, when I kept coming up with an answer fifty times smaller than what the back of the book had. Even though any book is prone to errors, I needed to make sure.

6. Jan 16, 2010

### CFDFEAGURU

Can you show how you go from kg/s^2 to N/m?

I follow the definition for a Newton but I don't see how to flip kg/s^2 to N/m without involving the gravitation constant.

I don't follow the wikipedia arcticle either. In my vibration text, "An Introduction to Mechanical Vibrations" by Robert Steidel 3rd Edition, the following is stated.

"For the total energy, E,

E = T + V

In this equation, T denotes kinetic energy due to the velocity of the mass of the system, and V is potential energy, either due to the configuration of this mass as measured from some arbitrary datum or due to the stress of the elastic member."

Let me know what the final outcome is. I apologize if I have helped you poorly.

Thanks
Matt

7. Jan 16, 2010

### ehild

Matt,

Newton= kg*m/s^2. N/m =(kg*m/s^2 )/m = kg/s^2.
Or backwards: kg/s^2= (kg*m/s^2) /m =N/m.

For a mass m, connected to a spring and performing simple harmonic motion, the total energy is 1/2 mv^2+kx^2, where x is the displacement from equilibrium, v is the velocity and k is the spring constant, and this total energy is conserved. At maximum displacement, v =0 and the magnitude of x is equal to the amplitude A. So the total energy is E=1/2 kA^2.

8. Jan 16, 2010

### Wellesley

No need to apologize....I'm sorry if I came across as ungrateful. I should have waited until the morning to reply, instead of writing my response at 11:30 PM.

Please, if you have any more thoughts on this problem, I would be grateful if you shared them.

Last edited: Jan 16, 2010
9. Jan 16, 2010

### CFDFEAGURU

No problem. I make mistakes everyday. Thats how I no that I am doing something. :rofl:

Thanks
Matt