How Accurate Are These Calculations for Simple Harmonic Motion?

Click For Summary
SUMMARY

The discussion focuses on the calculations related to simple harmonic motion involving a 0.5 kg block colliding with a massless pan attached to a spring with a spring constant of 32 N/m. The displacement function is accurately determined as x(t) = 0.25sin(8t), with a period T of 0.785 seconds. The kinetic energy of the mass 4.0 seconds post-collision is calculated to be 0.997 J, and the force exerted by the spring on the block at t = 1.2 seconds is -1.334 N, indicating the block is moving towards the equilibrium position. The importance of showing working steps for verification is emphasized.

PREREQUISITES
  • Understanding of simple harmonic motion principles
  • Knowledge of spring constant and its role in oscillations
  • Familiarity with kinetic energy calculations
  • Ability to analyze forces in oscillatory systems
NEXT STEPS
  • Study the derivation of displacement functions in simple harmonic motion
  • Learn about energy conservation in oscillatory systems
  • Explore the effects of damping on simple harmonic motion
  • Investigate the relationship between force and displacement in Hooke's Law
USEFUL FOR

Students of physics, educators teaching mechanics, and anyone interested in understanding the dynamics of oscillatory systems.

nyyfan0729
Messages
12
Reaction score
0
A block of mass 0.5kg moving on a horizontal frictionless surface at 2.0 m/s collides with and sticks to a massless pan attached to the end of a horizontal ideal spring whose spring constan is 32 N/m.
a) Determine the function for x(t), the displacement from equilibrium position as a function of time.
I got this to be x(t) = .25sin(8t)

b) What is the period T, of the subsequent oscillations?
I got the period to be .785 sec

c) What is the kinetic energy of the mass 4.0 sec after it collides with the spring?
I got this to be .997j

d) What force is exerted by the spring on the block at t=1.2 sec? Which way is the block moving? Explain
I got the force to be -1.334N

ARE MY ANSWERS RIGHT, I ALSO SAID THAT Xo=0m when using my equation.
 
Physics news on Phys.org
You have to show your working, it is much easier to check for systematic errors, you can't expect someone to work through your problems from scratch.
 

Similar threads

  • · Replies 51 ·
2
Replies
51
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
16
Views
2K
Replies
3
Views
994
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 17 ·
Replies
17
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K