Oscillations

JEE Main level, single correct, medium

158 of 194

Question 158 of 194

For a simple harmonic motion in a mass spring system shown, the surface is frictionless. When the mass of the block is 1kg1kg, the angular frequency is ω1{\omega }_{1}. When the mass block is 2kg2kg the angular frequency is ω2{\omega }_{2}. The ratio ω2ω1\dfrac{{\omega }_{2}}{{\omega }_{1}} is :

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Answer: (B) 12\dfrac{1}{\sqrt{2}}

In case of SHM in spring-block system, angular frequency is given by, ω=km\omega =\sqrt{\dfrac{k}{m}}.
Therefore,
ω2ω1=m1m2=12\dfrac{{\omega }_{2}}{{\omega }_{1}}=\sqrt{\dfrac{{m}_{1}}{{m}_{2}}}=\dfrac{1}{\sqrt{2}}

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