Simple Harmonic Motion w/ postion function

In summary: I appreciate your help. :)In summary, the given object is performing simple harmonic motion with an amplitude of 3.50m. Its period is 4 seconds and its frequency is 0.25 Hz. The maximum kinetic energy is 7.56 J and the maximum potential energy is also 7.56 J. The total mechanical energy of the system is 15.12 J.
  • #1
xxsteelxx
27
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 equation 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!
 
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  • #2
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
xxsteelxx said:
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 equation 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
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
Thank you very much! I realize that I need to review my energy concepts.
 

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. This means that the object will oscillate back and forth between two points with a constant frequency.

2. What is the position function for Simple Harmonic Motion?

The position function for SHM is given by x(t) = A cos(ωt + φ), where A is the amplitude, ω is the angular frequency, t is time, and φ is the phase constant. This function represents the displacement of the object from its equilibrium position at any given time.

3. What is the relationship between period and frequency in Simple Harmonic Motion?

The period (T) of SHM is the time it takes for one complete cycle of oscillation, while the frequency (f) is the number of cycles per unit time. They are inversely proportional, meaning that as the period increases, the frequency decreases, and vice versa. This relationship is given by T = 1/f or f = 1/T.

4. What is the role of the spring constant in Simple Harmonic Motion?

The spring constant (k) is a measure of the stiffness of the spring in SHM. It determines the strength of the restoring force and therefore, affects the frequency and period of the motion. A larger spring constant will result in a smaller period and higher frequency.

5. What are some real-life examples of Simple Harmonic Motion?

Some common examples of SHM include a swinging pendulum, a mass on a spring, and a vibrating guitar string. It can also be seen in the motion of particles in a liquid or gas, the motion of stars in a galaxy, and many other natural phenomena.

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