A mass ' m₁ ' is suspended from a spring of negligible mass. A spring is pulled slightly in downward direction and released, mass performs S.H.M. of period ' T₁ '. If the mass is increased by ' m₂ ', the time period becomes ' T₂ '. The ratio (m₂)/(m₁) is (a) (T₁²+T₂²)/(T₁²) (b) (T₁-T₂)/(T₁) (c) (T₂²-T₁²)/(T₁²) (d) (T₁²-T₂²)/(T₁²)

Answer: (c). T ∝ √(m): (T₂²)/(T₁²) = (m₁ + m₂)/(m₁) = 1 + (m₂)/(m₁). (m₂)/(m₁) = (T₂²)/(T₁²) - 1 = (T₂² - T₁²)/(T₁²) So the answer is (c).

A mass ' m₁ ' is suspended from a spring of negligible mass. A spring… | MHT CET 2021 (20 Sep, Shift 1)

Q67MHT CET 2021 · 20 SEP · SHIFT 1Medium

Spring Systems

A mass ' $m_1$ ' is suspended from a spring of negligible mass. A spring is pulled slightly in downward direction and released, mass performs S.H.M. of period ' $T_1$ '. If the mass is increased by ' $m_2$ ', the time period becomes ' $T_2$ '. The ratio $\frac{m_2}{m_1}$ is

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Answer: (c)

$T \propto \sqrt{m}$ : $\frac{T_2^2}{T_1^2} = \frac{m_1 + m_2}{m_1} = 1 + \frac{m_2}{m_1}$ .
$\frac{m_2}{m_1} = \frac{T_2^2}{T_1^2} - 1 = \frac{T_2^2 - T_1^2}{T_1^2}$
So the answer is (c).

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Spring Systems question from MHT CET 2021 (20 Sep, Shift 1)

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