Simple Harmonic Motion, masses, and springs.

Click For Summary

Homework Help Overview

The discussion revolves around a problem involving simple harmonic motion, specifically focusing on two blocks and a spring system on a frictionless surface. The original poster is tasked with finding the maximum amplitude of motion without slippage occurring between the blocks, given specific masses, spring constant, and coefficient of static friction.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to derive an expression for maximum amplitude using free body diagrams and equations of motion. Some participants question the correctness of the free body diagrams and the equations used, while others suggest differentiating the position function to analyze acceleration.

Discussion Status

Participants are actively engaging with the problem, with some providing guidance on checking units and re-evaluating steps in the calculations. There is a recognition of errors in the original poster's approach, but no consensus has been reached on the correct method yet.

Contextual Notes

There is mention of potential errors in the free body diagrams and the need to check units, indicating that assumptions about the setup may need to be revisited. The original poster expresses uncertainty about which equations to differentiate, highlighting a lack of clarity in the problem interpretation.

csnsc14320
Messages
57
Reaction score
1

Homework Statement


Two blocks (m = 1.22 kg and M = 8.73 kg) and a spring (k = 344 N/m) are arranged on a horizontal, frictionless surface, with m resting on top of M and the spring attached to a wall and M. The coefficient of static friction between the blocks is 0.42. Find the maximum possible amplitude of the simple harmonic motion is no slippage is to occur between the blocks.


Homework Equations





The Attempt at a Solution


So, after drawing free body diagrams for both blocks, I get:

m: [tex]\sum F = -\mu_s m g = m a[/tex] or just [tex]a = -\mu_s g[/tex]

M: [tex]-k x = (M + m) a[/tex] or just [tex]a = -\frac{M + m}{k x}[/tex]

Setting these two equations equal, I finally get:

[tex]x = \frac{M + m}{k \mu_S g}[/tex]

However, when I plug in the values, I get the wrong answer. I think something may be wrong with my FBD's?

Thanks in advance.
 
  • Like
Likes   Reactions: Sunav K Vidhyarthi
Physics news on Phys.org
The acceleration is a function of time. Diffentiate the position function twice with respect to time and find the maximum acceleration. This expression will give the acceleration as a function of omega and amplitude.
 
Wouldn't the acceleration be the greatest at the maximum displacement? That's why I thought it would be find to only consider the acceleration at the point of maximum amplitude, x.

Also, I have two equations for the acceleration? which would I differentiate? Or does it matter?
 
Your method is fine; you just made a simple error.
csnsc14320 said:
M: [tex]-k x = (M + m) a[/tex]
OK.
or just [tex]a = -\frac{M + m}{k x}[/tex]
Oops... redo this step. (You would have caught this if you had checked units.)
 
Doc Al said:
Your method is fine; you just made a simple error.

OK.

Oops... redo this step. (You would have caught this if you had checked units.)

Oops! I need to slow down a little bit when doing my simple algebra :-P

I re-did it, checked my units (came out to meters),plugged it, and got the right answer.

Thanks
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
1K
Replies
16
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 10 ·
Replies
10
Views
4K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
7
Views
1K
  • · Replies 5 ·
Replies
5
Views
40K
  • · Replies 3 ·
Replies
3
Views
7K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
9
Views
3K