Question #b7afd

Question

2637 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-03T09:52:56

The additional mass is $=0.389kg$

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

$=0.889-0.500=0.389kg$