For a simple harmonic motion in a mass spring system shown, the surface is frictionless. When the mass of the block is 1 kg, the angular frequency is ω₁. When the mass block is 2 kg the angular frequency is ω₂. The ratio ω₂/ω₁ is (a) 1/√(2) (b) 1/2 (c) 2 (d) √(2)

Answer: (a). Same spring, frictionless surface: ω = √(k/m), so ω ∝ 1/√(m). (ω₂)/(ω₁) = √((m₁)/(m₂)) = √(1/2) So the answer is (a) 1/√(2).

For a simple harmonic motion in a mass spring system shown, the… | JEE Main 2023 (30 Jan, Shift 2)

Q46JEE Main 2023 · 30 JAN · SHIFT 2Easy

Spring Systems

For a simple harmonic motion in a mass spring system shown, the surface is frictionless. When the mass of the block is $1\text{ kg}$ , the angular frequency is $\omega_1$ . When the mass block is $2\text{ kg}$ the angular frequency is $\omega_2$ . The ratio $\omega_2/\omega_1$ is

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Answer: (a)

Same spring, frictionless surface: $\omega = \sqrt{\frac{k}{m}}$ , so $\omega \propto \frac{1}{\sqrt{m}}$ .
$\frac{\omega_2}{\omega_1} = \sqrt{\frac{m_1}{m_2}} = \sqrt{\frac{1}{2}}$
So the answer is (a) $1/\sqrt{2}$ .

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