A particle at the end of a spring executes SHM with a period t_(1), while corresponding period for another spring is t_(2). If the period of oscillation when the two springs are connected in series is T, then prove that, t_(1)^(2)+t_(2)^(2)=T^(2).

A particle at the end of a spring executes SHM with a period t_(1), wh

1.	A particle at the end of a spring executes SHM with a period t_(1), while corresponding period for another spring is t_(2). If the period of oscillation when the two springs are connected in series is T, then prove that, t_(1)^(2)+t_(2)^(2)=T^(2).
Answer» Solution :LET the mass of the particle be m and SPRING constants of the SPRINGS be `k_(1)andk_(2)`. In first case, `t_(1)=2pisqrt(m/k_(1))or,t_(1)^(2)=4pi^(2)(m/k_(1))""...(1)` In second case, `t_(2)=2pisqrt(m/k_(2))or,t_(2)^(2)=4pi^(2)(m/k_(2))""...(2)` In series the equivalent spring CONSTANT is k, Then, `1/k=1/k_(1)+1/k_(2)or,k=(k_(1)k_(2))/(k_(1)+k_(2))` `therefore` Time PERIOD of the combination, `T=2pisqrt(m/k)=2pisqrt((m(k_(1)+k_(2)))/(k_(1)k_(2)))` or, `T^(2)=(4pi^(2)m(k_(1)+k_(2)))/(k_(1)k_(2))` adding (1) and (2) we get, `t_(1)^(2)+t_(2)^(2)=4pi^(2)(m/k_(1)+m/k_(2))=4pi^(2)m(1/k_(1)+1/k_(2))` `=4pi^(2)m((k_(1)+k_(2))/(k_(1)k_(2)))` `therefore" "t_(1)^(2)+t_(2)^(2)=T^(2)` (Proved).