What frequency of vibration is required?

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Homework Help Overview

The problem involves simple harmonic motion, specifically examining the relationship between acceleration, frequency, and displacement in a vibrating system. The original poster seeks to determine the frequency of vibration required given an acceleration amplitude and a vibration amplitude, along with related questions about maximum speed and time delay between points of maximum acceleration and maximum speed.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the equation for acceleration in simple harmonic motion and question the validity of the original poster's calculations. There are attempts to derive frequency from given parameters, with some participants questioning the signs in equations and the correctness of the formulas used. Others suggest alternative approaches and clarify the relationship between acceleration, velocity, and displacement.

Discussion Status

Multiple interpretations of the equations are being explored, with some participants offering guidance on the correct formulas and relationships. There is ongoing questioning about the assumptions made, particularly regarding the signs in the equations and the definitions of terms. The discussion is active, with participants seeking clarification and verification of their reasoning.

Contextual Notes

There is a noted confusion regarding the signs in the acceleration formula and the implications of the phase difference between maximum acceleration and maximum speed. Participants are also navigating the constraints of homework rules and the need for accurate formulas as provided in their materials.

  • #31
f=1/t
t= 1/f

= 1/ 7.11
t= 0.14 seconds...

is this the time delay...then?
 
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  • #32
NO that's the period. Do you know what that means? How much of the period goes by in pi/2 of the wavelength? That's what you need to get.
 
  • #33
not sure...
well i know speed= frequncy * wavelength
 
  • #34
No. OK, the period is the time for one wavelength. You have that now. So if pi/2 (or 90 degrees) of the wavelength goes by, how much time is that? Basically how much time for 1/4 of the wavelength? You don't need velocity for this.
 
  • #35
f=1/t
t= 1/f

= 1/ 7.11
t= 0.14 seconds

0.14/4= 0.035 sec
 
  • #36
imy786 said:
(iii)
x= Asin (wt)

v= Awcos (wt)
100/w= v

100= Aw^2 (wt)

A is a constant, what will this constant be?

would this be right equation for (iii)

Did you draw a wave? Starts at 0 (at t=0), goes up to A (at t=(1/2)*pi/w)), goes back down to 0 (at t=pi/w), goes down to -A (at t=(3/2)*pi/w) finally back up to 0 (t=2*pi/w). For the first part of iii) since a=-w^2*sin(wt), points of maximum acceleration will be when A*sin(wt) is a maximum in size. So this could be either when it is A or -A. From the graph we can see this happens at t=(1/2)*pi/w and t=(3/2)*pi/w. So the interval between two such events will be (fill in the answer here). As for maximum speed, v=A*cos(wt). If you play the same game you did with sin(wt) you will see this is A at t=0, -A at t=pi/w and +A again at 2*pi/w and these are the points when the SIZE of v is a maximum. So the interval between two such events is (fill in the answer here).
 
  • #37
What is the time delay between the points of maximum acceleration and
of maximum speed?

t=(1/6)*pi/w ...IS this correct?

(3/2)*pi/w- (1/2)*pi/w
 
  • #38
(3/2)*pi/w-(1/2)*pi/w is the difference between two points of maximum acceleration. But it's not (1/6)*pi/w.
 
  • #39
Dick,
I was under the impression that the OP is looking for the time difference between the occurrence of the maximum acceleration and the maximum speed in the wave, not the time between each maximum acceleration or each maximum speed (if I understand you correctly). But I can see that question is a little vague that way, so now I'm not sure.

imy786,
if you do what I think the question is asking, the answer you gave in post #35 seems OK to me.
But if the question is really asking what Dick is saying, then the answer will be different. You must decide which way is the one that your question is asking.
 
  • #40
Yeah, I agree. I was hoping imy786 would clarify which was wanted.
 
Last edited:

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