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 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. (T₁+T₂)/(T₁-T₂) = (2+1)/(2-1) = 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 2020 (19 Oct, Shift 2)

Q77MHT CET 2020 · 19 OCT · SHIFT 2Easy

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 $4\,\frac{\text{N}}{\text{m}}$ and $16\,\frac{\text{N}}{\text{m}}$ respectively. If $T_1$ and $T_2$ are period of oscillations of springs $A$ and $B$ respectively, then $\frac{T_1+T_2}{T_1-T_2}$ will be

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Answer: (c)

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

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Spring Systems question from MHT CET 2020 (19 Oct, Shift 2)

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