Harmonic Motion of Oscillating Particle

Click For Summary
A particle initially positioned at 0.280 m with a velocity of 0.200 m/s and an acceleration of -0.450 m/s² is analyzed under two scenarios: constant acceleration and simple harmonic motion (SHM). For the constant acceleration case, the position and velocity after 4.10 seconds are calculated to be -2.68 m and -1.65 m/s, respectively. In the SHM scenario, the angular frequency is related to the acceleration being proportional to the displacement from equilibrium, and the amplitude can be determined using conservation of energy principles. The phase constant must be calculated considering the two possible angles from the inverse trigonometric function. The final position and velocity after 4.10 seconds of oscillation need to be derived from the SHM equations provided.
MD_Programmer
Messages
10
Reaction score
0

Homework Statement


A particle moves along the x axis. It is moving initially at the position 0.280 m, moving with velocity 0.200 m/s and acceleration -0.450 m/s^2. Suppose it moves with constant acceleration for 4.10 s.


(a) Find the position of the particle after this time.

(b) Find its velocity at the end of this time interval.

We take the same particle and give it the same initial conditions as before. Instead of having a constant acceleration, it oscillates in simple harmonic motion for 4.10 s around the equilibrium position x = 0.
(c) Find the angular frequency of the oscillation. Hint: in SHM, a is proportional to x.

(d) Find the amplitude of the oscillation. Hint: use conservation of energy.

(e) Find its phase constant 0 if cosine is used for the equation of motion. Hint: when taking the inverse of a trig function, there are always two angles but your calculator will tell you only one and you must decide which of the two angles you need.

(f) Find its position after it oscillates for 4.10 s.

(g) Find its velocity at the end of this 4.10 s time interval.

Homework Equations


x(t) = A cos(wt + phi)
w = sqrt(k/m)
v = dx/dt = -w Asin(wt + phi)
Vmax = w A = sqrt(k/m) A
a = d^2x/dt^2 = -w^2 Acos(wt + phi)
T = 2pi sqrt(m/k)
f = 1/T
w = 2pi f



The Attempt at a Solution


found a to be -2.68 m
found b to be -1.65 m/s
can't figure out c-g please help :(
 
Physics news on Phys.org
found a to be -2.68 m
found b to be -1.65 m/s
can't figure out c-g please help :(
... please show your working and your reasoning behind each of these answers, and you best attempt so far for the last ones: how you think about these problems will help me guide you.

Guessing: for c. the acceleration is no longer constant.
The given number is the instantaneous acceleration at the given position.
(that help)

Also you are missing an equation - very important one from the definition of SHM.
Usually about the way the restoring force varies with position.
 
Last edited:
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 13 ·
Replies
13
Views
1K
Replies
1
Views
1K
Replies
13
Views
2K
Replies
9
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 13 ·
Replies
13
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
3
Views
815
Replies
16
Views
2K