Oscillations

JEE Main level, numerical answer, medium

154 of 194

Question 154 of 194

The time period of simple harmonic motion of mass MM in the given figure is παM5K\pi \sqrt{\dfrac{\alpha M}{5K}}, where the value of α\alpha is _______.

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Answer: 12

For series combination of the spring, equivalent spring constant is given by, 1keq=1k1+1k2\dfrac{1}{{k}_{eq}}=\dfrac{1}{{k}_{1}}+\dfrac{1}{{k}_{2}}.
Therefore, for combined system net spring constant keq=2k×k2k+k+k=5k3{k}_{eq}=\dfrac{2k\times k}{2k+k}+k=\dfrac{5k}{3}
Angular frequency of oscillation (ω)=keqm\left(\omega \right)=\sqrt{\dfrac{{k}_{eq}}{m}}
⇒(ω)=5k3m\Rightarrow \left(\omega \right)=\sqrt{\dfrac{5k}{3m}}
Period of oscillation (τ)=2πω=2π3m5k\left(\tau \right)=\dfrac{2\pi }{\omega }=2\pi \sqrt{\dfrac{3m}{5k}}
=π12m5k=\pi \sqrt{\dfrac{12m}{5k}}
Therefore, α=12\alpha =12.

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