1.

A mass is suspended separately by two different springs in successive order then time periods is t_(1) and t_(2)respectively. If it is connected by both spring as shown in figure then time period is to, the correct relation is

Answer»

`t_(0)^(2) = t_(1)^(2) + t_(2)^(2)`
`t_(0)^(-2) = t_(1)^(-2) + t_(2)^(-2)`
`t_(0)^(-1) = t_(1)^(-1) + t_(2)^(-1)`
`t_(0) = t_(1) + t_(2)`

Solution :The time period of a spring mass system is given by,
`T =2pi sqrt(m//k)`, where ks is the spring CONSTANT.
`therefore t_(1) = 2pi sqrt(m//k_(1))`..........(i)
and `t_(2) = 2pi sqrt(m//k_(2))`.........(II)

Now, when they are CONNECTED in parallel as shown in figure (a), the system can be replaced by a single spring of spring constant `k_(eff)= k_(1) + k_(2)`.
[Since, `mg = k_(1)x + k_(2)x = k_(eff)x`]
`therefore t_(0) =2pi sqrt(m//k_(eff)) = 2pisqrt(m//(k_(1)+k_(2)))`.........(iii)
From (i), `1/t_(1)^(2) = 1/(4pi^(2)) xx k_(1)/m`
From (ii), `1/t_(2)^(2) = 1/(4pi^(2)) xx k_(2)/m`...........(v)
From (iii), `1/t_(0)^(2) = 1/(4pi^(2)) xx (k_(1) + k_(2))/m`.............(vi)
ADDING equation (IV) and (v), we get
`=1/t_(1)^(2) + 1/t_(2)^(2) =1/(4pi^(2)m) (k_(1) + k_(2)) =1/(t_(0)^(2))`
`therefore t_(0)^(-2) =t_(1)^(-2) + t_(2)^(-2)`


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