You hang a mass m $=0\mathrm{.}22$ kg on a spring with spring constant $12$ N$/$m$.$  You pull the mass down a distance of $0\mathrm{.}04$ m from its equilibrium position and release it$.$ You measure the period to be  

 

 $0\mathrm{.}85$ s$.$ 

 A If you instead pull the mass down $0\mathrm{.}12$ m from its equilibrium position and release it$,$ what would the new period be$?$  Justify your answer 

  

 B If you use a spring whose spring constant is one$-$fourth the original spring constant, how would the new period compare to the original period?  

 

 < The new period will be the same. 

 

<  The new period will be longer. 

 

<  The new period will be shorter. 

 

 

 

  

 

Answer rating: 100% (QA)

I A New Period when the mass is pulled down by 012 m The period of oscillation for a massspring syst... View the full answer