Simple Harmonic Motion using total mechanical energy

In summary, the conversation discusses finding the amplitude of motion in a system with a 250 gram mass connected to a spring and executing simple harmonic motion with a period of 0.5 seconds and total mechanical energy of 0.50J. The equation ΔU = 1/2kx2 is used to solve for the amplitude, but the conversation also considers other equations such as kx=ma and d2x/dt2=-kx. Ultimately, it is determined that the solution for x will be in the form of x = F(t), where F is a recognizable function.
  • #1
StrawHat
33
0

Homework Statement


A 250 gram mass is connected to a spring and executes simple harmonic motion. The period of motion is 0.5 seconds and the total mechanical energy is 0.50J. What is the amplitude of motion?

Homework Equations


ΔU = 1/2kx2

The Attempt at a Solution


I get

1/2kx2 = 0.5J,

then I get

kx2 = 1.0J

Not sure where to go from here. I do have the answer from the answer key, but I have no idea how to actually get the answer. I think I'm supposed to integrate something, but I'm not sure how to incorporate the time value into any equations.
 
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  • #2
do you know another equation involving k that applies to your situation?
 
  • #3
kx=ma, perhaps? If so, should I use kx = m(dv/dt)? But then how will I obtain a value for velocity?
 
Last edited:
  • #4
what does the solution of the differential equation kx=ma look like? [Check your notes on simple harmonic motion, watch out for sign conventions]
 
  • #5
d/dt(kx)=d/dt(ma)
k(dx/dt)=m(da/dt)
kv(t)=m(da/dt)?
 
  • #6
a = dv/dt = d2x/dt2 would be a better route. da/dt is going in the wrong direction.

========


the solution will be of the form x = F(t) where F will be a function that you recognise.
 
Last edited:
  • #7
x=ma/k? I am totally lost...

I have ax=4. Am I on the right track?
 
  • #8
what do your notes say for how x varies in a system that is executing simple harmonic motion? There is something not quite right with your kx = ma. Not quite because it is normally expressed in an ever so slightly different way.
 
  • #9
F = -kx
W = Fd =∫Fnetdx
Wtotal = ΔK
ΔU = 1/2kx2
 
  • #10
If y = sin(t) what is dy/dt, what about d2y/dt2
 
  • #11
dy/dt=cos(t), d2y/dt2=-sin(t)?

Thank you for your help thus far, but it's 4AM over here in the EST timezone, so I must go to bed. I will check back on this thread in five hours or so.
 
  • #12
so how are y and d2y/dt2 related? is there anything that you have posted so far that looks similar?
 
  • #13
m(dv/dt)=k(dx/dt)
m(dv/dt)=kv?
 

1. What is Simple Harmonic Motion?

Simple Harmonic Motion (SHM) is a type of periodic motion where an object oscillates back and forth around an equilibrium point under the influence of a restoring force that is directly proportional to the displacement from the equilibrium point.

2. How is total mechanical energy related to Simple Harmonic Motion?

Total mechanical energy in SHM is the sum of kinetic energy and potential energy. As the object oscillates, it constantly exchanges kinetic and potential energy, but the total mechanical energy remains constant.

3. What is the equation for total mechanical energy in Simple Harmonic Motion?

The equation for total mechanical energy is E = 1/2kA^2, where k is the spring constant and A is the amplitude (maximum displacement) of the oscillation.

4. How does the amplitude affect the total mechanical energy in Simple Harmonic Motion?

The amplitude directly affects the total mechanical energy in SHM. As the amplitude increases, the total mechanical energy also increases. This is because the potential energy is directly proportional to the square of the amplitude.

5. What is the relationship between frequency and total mechanical energy in Simple Harmonic Motion?

Frequency and total mechanical energy in SHM are inversely proportional. As the frequency increases, the total mechanical energy decreases. This is because frequency is directly proportional to the square root of the spring constant, and the total mechanical energy is directly proportional to the square of the spring constant.

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