Simple Harmonic Motion of a wave

Click For Summary
SUMMARY

The discussion focuses on the analysis of simple harmonic motion (SHM) as applied to a cheerleader waving a pom-pom with an amplitude of 18.0 cm and a frequency of 0.850 Hz. Key calculations include the maximum acceleration of 5.13 m/s² and maximum velocity of 0.961 m/s, derived using the formulas abs(a_max) = ω²A and abs(v_max) = ωA, where ω = 5.34 rad/s. The discussion also addresses the acceleration and speed at x = +9.0 cm, yielding -4.45 m/s² and 0.832 m/s, respectively. Furthermore, it clarifies which quantities can be determined using the energy approach, concluding that maximum values can be calculated through energy conservation, while specific instantaneous values and time cannot.

PREREQUISITES
  • Understanding of simple harmonic motion (SHM)
  • Familiarity with angular frequency (ω) and its calculation
  • Knowledge of energy conservation principles in physics
  • Ability to apply kinematic equations in SHM
NEXT STEPS
  • Study the derivation of angular frequency (ω) in simple harmonic motion
  • Learn about the relationship between kinetic and potential energy in SHM
  • Explore the mathematical modeling of SHM using differential equations
  • Investigate the implications of phase and amplitude on SHM behavior
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and wave motion, as well as educators seeking to enhance their understanding of simple harmonic motion concepts.

iwonde
Messages
30
Reaction score
0
[SOLVED] Simple Harmonic Motion

Homework Statement


A cheerleader waves her pom-pom in simple harmonic motion with an amplitude of 18.0 cm and a frequency of 0.850 Hz. Find (a) the maximum magnitude of the acceleration and of the velocity; (b) the acceleration and speed when the pom-pom's coordinate is x = +9.0 cm; (c) the time required to move from the equilibrium position directly to a point 12.0 cm away. (d) Which of the quantities asked for in parts (a), (b), and (c) can be found using the energy approach, and which ones cannot? Explain.


Homework Equations


abs(a_max)= ω^2A, A = amplitude
abs(v_max)=ωA
ω=2πf, f=frequency
a = -ω^2 sqrt(A^2 - x^2)



The Attempt at a Solution


(a) ω=2π(0.85)= 5.34 rad/s
abs(a_max)= (5.34)^2(0.18)= 5.13m/s^2
abs(v_max)= (5.34)(0.18) = 0.961 m/s

(b) a = -(5.34)^2 sqrt((0.18)^2 - (0.09)^2) = -4.45m/s^2
speed = abs(v) = (5.34) sqrt((0.18)^2 - (0.09)^2) = 0.832 m/s

I don't know how to approach (c) and (d).
 
Physics news on Phys.org
Maybe this will do for the (c) part:
x=As in\omega t.

(d) part seems easy. Just try calculating every part using the law of conservation:
E=E_{k}+E_{p}=E_{k(max)}=E_{p(max)}.
 
Last edited:
for d) I'm thinking that part (a) can be found by using the energy approach because K_max = (1/2)m(v_max)^2, from v_max, you can find a_max; part (b) cannot be found by using the energy method because energy doesn't deal with what happens in the beginning and end; part (c) cannot be found using the energy approach because the time variable doesn't occur in the energy equations.

Is that right?
 
The velocity can be found at any given point through conservation of energy.
 
I'm still a little confused about the energy approach in part (d).
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
3K
Replies
1
Views
6K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 13 ·
Replies
13
Views
1K
Replies
16
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K