Maximum speed of a mass on a spring

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SUMMARY

The maximum speed (Vmax) of a mass on a spring is definitively expressed by the formula Vmax = 2πfA, where f represents frequency and A denotes amplitude. The discussion highlights the relationship between kinetic energy and potential energy at maximum speed, establishing that all energy is kinetic at this point. Key equations include T = 2π√(m/k) and f = 1/2π√(k/m), which are essential for deriving the maximum speed formula. The participant successfully eliminates variables to arrive at the correct expression for Vmax.

PREREQUISITES
  • Understanding of harmonic motion principles
  • Familiarity with kinetic and potential energy equations
  • Knowledge of the spring constant (k) and mass (m) relationship
  • Basic grasp of trigonometric functions and their applications in physics
NEXT STEPS
  • Study the derivation of the spring constant (k) in relation to mass (m) and frequency (f)
  • Explore the implications of energy conservation in oscillatory systems
  • Learn about the effects of damping on harmonic motion
  • Investigate real-world applications of spring dynamics in engineering
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Students studying physics, particularly those focusing on mechanics and oscillatory motion, as well as educators seeking to clarify concepts related to springs and harmonic motion.

the_morbidus
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Homework Statement



Prove that the maximum speed (Vmax) of a mass on a spring is given by 2PifA

f=frequency
A=amplitude


Homework Equations



ac=v^2/r

T=2Pir/v

ac=4 Pi^2 r / T^2

T=2Pi sqrt(r/ac) , T=2Pi sqrt(A/ac)

-x/a=m/k

T=2Pi sqrt(-x/ac) , T=2Pi sqrt(m/k)

f=1/2Pi sqrt(k/m) , f=1/2Pi sqrt(a/-x)

The Attempt at a Solution



Ok so after much trial and failure and producing much gibberish full of variables... This is what i know so far that is right, is the mass is going at max speed then the total energy which is made of potential energy plus kinetic energy, is all made of just kinetic energy because potential energy should equal 0 at max speed, well i think so...

Et=Ek
kA^2=mv^2

and anything after that doesn't make sense whatever route i try.

I am taking a correspondence course by ILC ( the independent learning centre at Ontario) , and its been 6 years since i graduated from high school so my grade 12 physics its not very good.

Thank you for all the help !
 
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the_morbidus said:
Ok so after much trial and failure and producing much gibberish full of variables... This is what i know so far that is right, is the mass is going at max speed then the total energy which is made of potential energy plus kinetic energy, is all made of just kinetic energy because potential energy should equal 0 at max speed, well i think so...

Et=Ek
kA^2=mv^2
Good thinking.

and anything after that doesn't make sense whatever route i try.
Hint: Try eliminating k/m using one of your equations. (First rewrite this equation in terms of k/m.)

For some reason, your list of relevant equations has stuff about centripetal acceleration. That's not relevant here.
 
o the thing about the centripetal equations is that the lesson was using them for explaining on how to obtain certain variations with T and f equations, so i tried everything but to no avail.

Dude i could kiss you! well metaphorically speaking hahaha, here is my work and well i hope this can be used for other people.

Et=Ek

kA^2=mv^2

k/m=v^2/A^2

sqrt (k/m) = v/A

v=A sqrt(v/A) i pause here and i work on another equation to insert on this one

T=2Pi sqrt(m/k)

(T/2Pi)^2=m/k

(2Pi/T)^2=k/m

2Pi/T=sqrt(k/m) now i insert on the previous equation and i get this

v=A (2Pi/T)

v= A2Pi/ T

v=A 2Pi f

got the result! thanks for the hint!
 

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