In figure (A), mass ‘2 m’ is fixed on mass ‘m’ which is attached to two spring of spring constant k. In figure (B), mass ‘m’ is attached to two springs of spring constant ‘k’ and ‘2k’. If mass ‘m’ in (A) and in (B) are displaced by distance ‘x’ horizontally and then released, then time period T₁ and T₂ corresponding to (A) and (B) respectively follow the relation.

Option 1 -

$\frac{T_{1}}{T_{2}} = \frac{3}{\sqrt[]{2}}$

Option 2 -

$\frac{T_{1}}{T_{2}} = \sqrt[]{\frac{3}{2}}$

Option 3 -

$\frac{T_{1}}{T_{2}} = \sqrt[]{\frac{2}{3}}$

Option 4 -

$\frac{T_{1}}{T_{2}} = \frac{\sqrt[]{2}}{3}$

2 Views | Posted 2 months ago

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1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

2 months ago

Correct Option - 1

Detailed Solution:

$T_{1} = 2 π \sqrt[]{\frac{3 m}{2 k}}$

$T_{2} = 2 π \sqrt[]{\frac{m}{3 k}}$

$\frac{T_{1}}{T_{2}} = \frac{2 π \sqrt[]{\frac{3 m}{2 k}}}{2 π \sqrt[]{\frac{m}{3 k}}} = \frac{3}{\sqrt[]{2}}$

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