Simple Harmonic Motion w/ postion function

  • Thread starter xxsteelxx
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  • #1
xxsteelxx
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The problem statement, with all known data/variables.
A 500 g object is moving a horizontal frictionless surface. Its displacement from the origin is given by the equation: [tex]x(t) = (3.50m)\sin{[(\frac{\pi}{2})t + \frac{5\pi}{4}]}[/tex].

a)what kind of motion is this?

b)what is the amplitude of this motion?

c)what is the period of this motion?

d)what is the frequency of this motion?

e)what is the linear velocity of this motion?

f)what is the linear acceleration of this motion?

g)what is the maximun kinetic energy of this system?

h)What is the maximum potential energy of this system?

i)What is the total mechanical energy of this system?

Given :
postion function x(t)
Amplitude: A=3.50m
mass = 500g= .500kg
no friction is present

Homework Equations


[tex]x'(t)=v(t)[/tex]
[tex]v'(t)=a(t)[/tex]
[tex]KE(t)= \frac{1}{2} m{[v(t)]}^2[/tex]
[tex]PE(t)= \frac{1}{2} k{[x(t)]}^2[/tex]
[tex]E_{mech} = \frac{1}{2} k{A}^2[/tex]
[tex]T= \frac{2\pi}{\omega}[/tex]
[tex]x(t)= A\sin{[ t\omega + \phi]}[/tex]
[tex]f=\frac{1}{T}[/tex]
[tex]\omega = \sqrt{\frac{k}{m}[/tex]

The Attempt at a Solution



a) since the postion function is sinusoidal, this reflects simple harmonic motion.
b) Given: A=3.50m (sinusoidal function)
c)[tex]T= \frac{2\pi}{\omega}[/tex]
[tex]T= \frac{2\pi}{\frac{\pi}{2}}[/tex]
[tex]T=4[/tex]

d) linear frequency [tex]f=\frac{1}{T}[/tex]
[tex]f=\frac{1}{4}[/tex]

e)differentiating x(t) gives [tex]v(t)= 3.50(\frac{\pi}{2})\cos{(\frac{\pi}{2} t + \frac{5\pi}{4})}[/tex]
f)differentiating v(t) gives [tex]a(t)= -3.50(\frac{{\pi}^2}{4})sin{(\frac{\pi}{2} t + \frac{5\pi}{4})}[/tex]
g) using the kinetic energy formula gives [tex]KE(t)= \frac{1}{2} (.500kg){[3.50(\frac{\pi}{2}) \cos(\frac{\pi}{2} t + \frac{5\pi}{4})]}^2[/tex]
To find max KE, we to find the time that velocity is greatest. Using velocity graph, we see max values at t=0,4: Therefore v is greatest at t=4. Plugging this in KE(t), we get KE(4)=3.778 Joules

h)First solve for k, [tex]T= \frac{2\pi}{\sqrt{\frac{k}{m}}}[/tex]
I get k= .20264, and I get stuck here.

i) Using total mech energy eqution I obtain total energy to be 1.24117 J

I sense that something is not right with parts (g),(h), and (i). And I am unsure if parts (a-f) are correct. Thanks in advance!
 

Answers and Replies

  • #2
ehild
Homework Helper
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The body performs a simple harmonic motion. At what position is the kinetic energy maximum? What is the potential energy there? Where is the potential energy maximum? What is the KE there? If the maximum kinetic energy is E(max), what is the total energy?

ehild
 
  • #3
ManyNames
136
0
The problem statement, with all known data/variables.
A 500 g object is moving a horizontal frictionless surface. Its displacement from the origin is given by the equation: [tex]x(t) = (3.50m)\sin{[(\frac{\pi}{2})t + \frac{5\pi}{4}]}[/tex].

a)what kind of motion is this?

b)what is the amplitude of this motion?

c)what is the period of this motion?

d)what is the frequency of this motion?

e)what is the linear velocity of this motion?

f)what is the linear acceleration of this motion?

g)what is the maximun kinetic energy of this system?

h)What is the maximum potential energy of this system?

i)What is the total mechanical energy of this system?

Given :
postion function x(t)
Amplitude: A=3.50m
mass = 500g= .500kg
no friction is present

Homework Equations


[tex]x'(t)=v(t)[/tex]
[tex]v'(t)=a(t)[/tex]
[tex]KE(t)= \frac{1}{2} m{[v(t)]}^2[/tex]
[tex]PE(t)= \frac{1}{2} k{[x(t)]}^2[/tex]
[tex]E_{mech} = \frac{1}{2} k{A}^2[/tex]
[tex]T= \frac{2\pi}{\omega}[/tex]
[tex]x(t)= A\sin{[ t\omega + \phi]}[/tex]
[tex]f=\frac{1}{T}[/tex]
[tex]\omega = \sqrt{\frac{k}{m}[/tex]

The Attempt at a Solution



a) since the postion function is sinusoidal, this reflects simple harmonic motion.
b) Given: A=3.50m (sinusoidal function)
c)[tex]T= \frac{2\pi}{\omega}[/tex]
[tex]T= \frac{2\pi}{\frac{\pi}{2}}[/tex]
[tex]T=4[/tex]

d) linear frequency [tex]f=\frac{1}{T}[/tex]
[tex]f=\frac{1}{4}[/tex]

e)differentiating x(t) gives [tex]v(t)= 3.50(\frac{\pi}{2})\cos{(\frac{\pi}{2} t + \frac{5\pi}{4})}[/tex]
f)differentiating v(t) gives [tex]a(t)= -3.50(\frac{{\pi}^2}{4})sin{(\frac{\pi}{2} t + \frac{5\pi}{4})}[/tex]
g) using the kinetic energy formula gives [tex]KE(t)= \frac{1}{2} (.500kg){[3.50(\frac{\pi}{2}) \cos(\frac{\pi}{2} t + \frac{5\pi}{4})]}^2[/tex]
To find max KE, we to find the time that velocity is greatest. Using velocity graph, we see max values at t=0,4: Therefore v is greatest at t=4. Plugging this in KE(t), we get KE(4)=3.778 Joules

h)First solve for k, [tex]T= \frac{2\pi}{\sqrt{\frac{k}{m}}}[/tex]
I get k= .20264, and I get stuck here.

i) Using total mech energy eqution I obtain total energy to be 1.24117 J

I sense that something is not right with parts (g),(h), and (i). And I am unsure if parts (a-f) are correct. Thanks in advance!

There are quite a few things to be mentioned, but i can't for now. I will return later and help you conceptualize this mathematically. :)
 
  • #4
Repainted
35
0
g) You are right to say that the KE is at a maximum when the velocity is at a maximum(in magnitude). Hence the angle in the cosine function of v(t) must be an integer multiple of pi. Which leads to t = 1.5(I don't know how you got t = 4?), or if you know your trig, you can simply conclude that the maximum value of cosine is = 1, hence the maximum velocity is simply (3.50)(pi/2) = 5.5m/s, maximum KE would then be (1/2)0.5(5.5)^2 = 7.56J.

h) Well, since energy is conserved, the maximum potential energy would simply be the same as the maximum KE.

i) Same here, the total mechanical energy is simply the same as the maximum amount of KE.
 
  • #5
xxsteelxx
27
0
Thank you very much! I realize that I need to review my energy concepts.
 

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