A body of mass 64 g is made to oscillate turn by turn on two different springs A and B. Spring A and B has force constant 4 N/m and 16 N/m respectively. If T₁ and T₂ are the period of oscillations of springs A and B respectively, then (T₁+T₂)/(T₁-T₂) will be (a) 1:2 (b) 1:3 (c) 3:1 (d) 2:1

Answer: (c). T ∝ 1/√(k) ⇒ (T₁)/(T₂) = √(16/4) = 2, so T₁ = 2T₂. (T₁+T₂)/(T₁-T₂) = (3T₂)/(T₂) = 3. So the answer is (c) 3:1.

A body of mass 64 g is made to oscillate turn by turn on two… | MHT CET 2026 (15 May, Shift 1)

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Spring Systems

A body of mass $64$ g is made to oscillate turn by turn on two different springs A and B. Spring A and B has force constant $4$ N/m and $16$ N/m respectively. If $T_1$ and $T_2$ are the period of oscillations of springs A and B respectively, then $(T_1+T_2)/(T_1-T_2)$ will be

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Answer: (c)

$T \propto \frac{1}{\sqrt{k}} \Rightarrow \frac{T_1}{T_2} = \sqrt{\frac{16}{4}} = 2$ , so $T_1 = 2T_2$ .
$\frac{T_1+T_2}{T_1-T_2} = \frac{3T_2}{T_2} = 3$ .
So the answer is (c) $3:1$ .

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