Question

A block of mass m is suspended separately by two different springs have time period t1 and t2. if same mass is connected to series combination of both springs, then its time period is t. then

Answer 1

Let spring constant of two different springs are k₁ and k₂ respectively.
We know, formula of time period of spring is given by
T = 2π√{m/k} , where m is mass and k is spring constant.

Now, in case of 1 :- time period , t₁ = 2π√{m/k₁}
squaring both sides,
t₁² = 4π²m/k₁ ⇒k₁ = 4π²m/t₁² --------(1)

Similarly in case of 2 :- time period ,t₂ = 2π√{m/k₂}
squaring both sides,
t₂² = 4π²m/k₂ ⇒k₂ = 4π²m/t₂² -------(2)

When both the given spring are connected in series combination then,
Equivalent spring constant , k = k₁k₂/(k₁ +k₂)
= {4π²m/t₁²}{4π²m/t₂²}/{4π²m}(1/t₁² + 1/t₂²)
= 4π²m/(t₁² + t₂²)
So, Time period = 2π√{m/k}
= 2π√{m/4π²m/(t₁² + t₂²)}
= √(t₁² + t₂²)