Calculating the maximum velocity & acceleration of a spring mass

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Homework Help Overview

The discussion revolves around a spring-mass system where a mass of 0.3 kg is suspended from a spring with a stiffness of 200 N/m. The original poster seeks to calculate various parameters related to the vibration of the mass after it is displaced from its equilibrium position.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate the frequency of vibration using the angular frequency formula but expresses confusion about applying the equations for maximum velocity and acceleration. Participants discuss the relationship between the equations and the conditions for maximum values.

Discussion Status

Some participants provide clarifications on the equations for velocity and acceleration, noting that maximum values occur when certain trigonometric functions reach their peak. There is an ongoing exploration of how to interpret these equations and apply them correctly.

Contextual Notes

The original poster mentions not asking for direct answers but seeks guidance on understanding the process. There is a focus on ensuring clarity around the use of trigonometric functions in the context of the problem.

hopkid
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Homework Statement



A mass of 0.3kg is suspended from a spring of stiffness 200Nm. If the mass is displaced by 10mm from it's equilibrium position and released, for the resulting vibration calculate:

a) the frequency of vibration
b) the maximum velocity of the mass during the vibration
c) the maximum acceleration of the mass during the vibration
d) the mass required to produce double the maximum velocity calculated using the same spring and initial deflection.

Homework Equations



w=√k/m
f=w/2Л
x=Asin(wt+Ø)
v=Aw cos(wt+Ø)
a=Aw² sin(wt+Ø)

The Attempt at a Solution



I have used w=√k/m to get w=√200/0.3 w=25.81 rads-1

The used f=w/2Л to get f=25.81/2x3.142 f=4.108 Hz

So i am ok working out the frequency of the vibration but i do not understand how to use the remaining equations to get the next answers, I'm not following the process. I'm not asking for answers here, just for someone to help me understand what i need to do to use these and achieve my answers.

Can anyone help me? Thank you in advance for any responses.

Daniel
 
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v= A \omega \cdot cos(\omega t)
A and omega are constants, so v is at it's max when cos(\omega t) is at it's max
 
Hi Daniel! :smile:
hopkid said:
x=Asin(wt+Ø)
v=Aw cos(wt+Ø)
a=Aw² sin(wt+Ø)

I don't understand what you're not getting about this …

dumbperson :smile: is correct ……

you have the equations for v and a …

what is the difficulty? :confused:
 
Thanks for the reply guys...

So, taking v=Aw cos(wt+Ø)

I get v = 0.01 x 25.81 cos(25.81t+Ø)
v= 0.2581 cos(25.81t+Ø)

So multiplying Aw (0.01 x 25.81) gets me 0.2581.

What happens with the second part of the equation cos(25.81t+Ø)?

That's leaving me a little confused.
 
The cos(and sin) of an angle can vary from a minimum value of zero to a maximum value of one.
It follows that v is a maximum when cos(wt+ phi) has its maximum value.
Similar reasoning can be used to find a max.
 
Thanks, so is 0.2581 mm/s-1 my final answer for vmax or do i need to work out cos(wt+ phi). If so how do i work out t and phi?
 
There is no need to do anything else with the angle other than take the maximum value of cos as being equal to one.In other words vmax=Aw.I haven't checked your numbers but you have presented your units incorrectly.
 
Many thanks for your patience :) Your help is much appreciated!
 

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