A period of oscillation of a mass m kg is suspended by an ideal spring is 1.5s. If an additional mass 3kg be suspended, the time period is increased by 1s. Find the value of m.

Explanation

In first case, `T_(1)=1.5s, m=m kg`
As `T=2pisqrt((m)/(k))So 1.5=2pisqrt((m)/(k))` …(i)
In second case, `T_(2)=1.5+1 =2.5s`,
mass`=(m+3)kg`
Then `2.5=2pisqrt((m+3)/(k))` …(ii)
Dividing (ii) by (i) , we get
`(2.5)/(1.5)=(sqrt(m+3))/(sqrt(3))or (5^(2))/(3^(2))=(m+3)/(m)orm=1.69kg`