A body of mass M, executes vertical SHM with periods t1

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A body of mass M, executes vertical SHM with periods t₁ and t₂, when separately attached to spring A and spring B respectively. The period of SHM, when the body executes SHM, as shown in the figure is t₀. Then

1. t₀^–1 = t₁^–1 + t₂^–1
2. t₀ = t₁ + t₂
3. t₀² = t₁² + t₂²
4. t₀^–2 = t₁^–2 + t₂^–2

Correct Option: D

t = 2π
		m
		k

⇒ k = Const . t^-2
Here the springs are joined in parallel. So
k₀ = k₁ - k₂
where k₀ is resultant force constant
or , Const . t₀^-2 = Const . t₁^-2 + Const . t₂^-2
or , t₀² = t₁² + t₂²