In the given figure, a body of mass $M$ is held between two massless springs, on a smooth inclined plane.

Question:

In the given figure, a body of mass $M$ is held between two massless springs, on a smooth inclined plane. The free ends of the springs are attached to firm supports. If each spring has spring constant $\mathrm{k}$, the frequency of oscillation of given body is :

  1. (1)

    $\frac{1}{2 \pi} \sqrt{\frac{2 k}{\operatorname{Mg} \sin \alpha}}$

  2. (2) $\frac{1}{2 \pi} \sqrt{\frac{\mathrm{k}}{\mathrm{Mg} \sin \alpha}}$

  3. (3) $\frac{1}{2 \pi} \sqrt{\frac{2 \mathrm{k}}{\mathrm{M}}}$

  4. (4) $\frac{1}{2 \pi} \sqrt{\frac{\mathrm{k}}{2 \mathrm{M}}}$


Correct Option: 1

Solution:

(1)

Equivalent $\mathrm{K}=\mathrm{K}+\mathrm{K}=2 \mathrm{~K}$

Now, $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{K}_{\mathrm{mq}}}}$

$\Rightarrow \mathrm{T}=2 \pi \sqrt{\frac{\mathrm{m}}{2 \mathrm{k}}}$

$\therefore \mathrm{f}=\frac{1}{2 \pi} \sqrt{\frac{2 \mathrm{k}}{\mathrm{m}}}$

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