Scaling factor of a mass spring system

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Homework Help Overview

The discussion revolves around a mass-spring system where the original poster seeks to determine the additional mass required to lower the resonant frequency from 15Hz to 10Hz. The problem involves understanding the relationship between mass, spring constant, and frequency.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to rearrange equations to isolate the variable representing the additional mass. Some participants question the approach of moving constants to one side of the equation and emphasize that the spring constant remains unchanged. Others note a misunderstanding regarding the relationship between period and frequency.

Discussion Status

Participants are exploring different methods to manipulate the equations and clarify misunderstandings. Some guidance has been offered regarding the nature of the spring constant and the relationship between frequency and period. There appears to be a productive exchange of ideas, though no consensus has been reached on the final solution.

Contextual Notes

There is a mention of a potential mistake in interpreting the relationship between frequency and period, as well as the need to consider the spring constant as a fixed property in the context of the problem.

mrcotton
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Homework Statement



A mass spring system carries a mass of 0.4kg. When the point of suspension is made to vibrate verticaly at a frequency of 15Hz resonance occurs. What mass should be added to the 0.4kg in order to reduce the resonant frequency to 10Hz.

a) 0.20kg
b) 0.40kg
c) 0.50kg
d) 0.60kg

Homework Equations



T=2∏√(m/k)
15=2∏√(0.4/k)

The Attempt at a Solution



15=2∏√(0.4/k)
10=2∏√(0.4+x/k)

Do I rearrange these so I have all the constants on the rhs then equate the lhs and solve for x if that makes sense?
Any help appreciated
 
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It wouldn't really make sense to try to 'put all the constants on the RHS of the equation' - everything in the first equation is a constant and we know that 'x', the only variable in the second equation, cannot be zero.

Consider, instead, that 'k', as a property of the spring, will not change between the two scenarios.

EDIT: Oh, and 'T' represents the period of oscillation rather than the frequency.
 
JayneDoe said:
It wouldn't really make sense to try to 'put all the constants on the RHS of the equation' - everything in the first equation is a constant and we know that 'x', the only variable in the second equation, cannot be zero.

Consider, instead, that 'k', as a property of the spring, will not change between the two scenarios.

EDIT: Oh, and 'T' represents the period of oscillation rather than the frequency.

I just realized the T=1/f mistake
Thanks for the help
 
0.06=2∏√(0.4/k)
0.1=2∏√(0.4+x/k)

So I still can't seem to get 0.5 out of the rearrangement and equation of the two formula.
 
I'm able to get x=0.5kg by that method, actually. If you leave '1/15' on the LHS of the first equation rather than approximating it as '0.06', you should find the same.
 
Thanks for the help,
I love the techniques of physics.
I got lost in the algebra

shm_zpsfa23b8c6.jpg
 

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