Question #67b4e

1 Answer
Jan 16, 2017

hyperphysics.phy-astr.gsu.edu

Let us consider that a sphere of mass #m# is hanged through spring of negligible mass from a support and the spring posseses spring constant #k#

If the spring is elongated by length #x# from its equilibrium position, then by Hooke's law the force exerted by the spring will be given by

#F_"spring"=-kx......[1]#

Now if the mass m executes SHM of ampltude #A# here then displacement #x# of mass #m# from equilibrium position will be related with time as follows

#x=Asin(omegat)......[2]#, where #omega# represents the angular velocity of refference point associated with the SHM.

Differentiating equation [2] w.r to t we get velocity #v#

#v=(dx)/(dt)=Aomegacos(omegat)=Awsqrt(1-sin^2(omegat))#
#=>v=omegasqrt(A^2-x^2).... [3]#

Differentiating [2] twice w.r to t we get acceleration (#a#) on mass #m#

#a=(d^2x)/(dt^2)=-Aomega^2sin(omegat)=-omega^2x....[4]#

So by Newtons law the restoring force on mass m when its displacement is #x#, will be given by

#ma=-momega^2x......[5]#

This force should be same as that obtained by Hooke's law earlier. So comparing [1] and [5] we get

#F_"spring"=-kx=ma=-omega^2x#

#=>omega^2=k/m#

#=>omega=sqrt(k/m)....[6]#

By equation [4] the maximum acceleration of the mass will be when #x=A#

So #a_"max"=omega^2A=k/mxxA#

Inserting #m=200g=0.2kg#

#k=400N/m# and #A=10cm=0.1m#

we get #a_"max"=400/0.2xx0.1=200"m/"s^2#

Putting #x=0# in [3] we get maximum velocity

#v_"max"=Aomega=Axxsqrt(k/m)#

#=0.1xxsqrt(400/0.2)"m/s"=sqrt20"m/s"#