Question #b7afd

1 Answer
Apr 3, 2017

The additional mass is #=0.389kg#

Explanation:

The period of oscillation of a spring is given by

#T=2pisqrt(m/k)#

The peroid is #T=1.5s#

The mass is #m=0.500kg#

and #k# is the spring constant

Therefore,

#m/k=T^2/(4pi^2)#

#k=(4pi^2)/T^2*m#

#k=(4pi^2)/(1.5^2)*0.5#

The new peroid is

#T_1=2pisqrt(M/k)#

#2=2pisqrt(M/((4pi^2)/(1.5^2)*0.5))#

#1=pi^2*M/((4pi^2)/(1.5^2)*0.5)#

#M=4*0.5/(1.5^2)#

#M=2/1.5^5=0.889kg#

The additional mass is

#=0.889-0.500=0.389kg#