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# A small block of mass ${\text{m}}$ is connected to a massless rod, which in terms attached in a spring of force constant $k$ , as shown in fig. If the block is displaced down slightly, and left free, then the time period of oscillation is?(A) $2\pi \sqrt {\dfrac{m}{k}}$ (B) $2\pi \left( {\dfrac{a}{b}} \right)\sqrt {\dfrac{m}{k}}$ (C) $2\pi \left( {\dfrac{b}{a}} \right)\sqrt {\dfrac{m}{k}}$ (D) $2\pi \sqrt {\dfrac{k}{m}}$

Last updated date: 20th Jun 2024
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Hint: This could be simply solved by breaking the diagrams into simple free body diagrams of both the blocks. Then we need to apply the frequency formula.

Formula used: Here, we will use the frequency formula:
${\omega _n} = \dfrac{g}{\vartriangle }$
Here, ${\omega _n}$ is the frequency
$\vartriangle$ is the deflection in the spring
Also,
${\omega _n} = \dfrac{{2\pi }}{T}$
Here, $T$ is the time period.

Complete step by step solution:
We will assume $F$ force acting at the left side on stretching of the spring due to the mass ${\text{m}}$ .
We will start by balancing the forces to find spring force,
$F \times k = mg$
$\Rightarrow F = \dfrac{{mg}}{k}$
Let us assume the deflection in spring be $k$
$x = \dfrac{b}{a}$
Deflection on mass ${\text{m}}$ :
$\vartriangle {\text{ = }}\dfrac{{\text{b}}}{{\text{a}}}\left( {{\text{deflection of spring}}} \right)$
$\Rightarrow \vartriangle {\text{ = }}\dfrac{{\text{b}}}{{\text{a}}}\left( {\dfrac{{mg}}{k}} \right) \times \dfrac{b}{a} = \dfrac{{mg{b^2}}}{{k{a^2}}}$
Also,
${\omega _n} = \dfrac{g}{\vartriangle } = \sqrt {\dfrac{{kg}}{{mg{{\left( {\dfrac{b}{a}} \right)}^2}}}}$
Equating the formula,
${\omega _n} = \dfrac{{2\pi }}{T}$
$\Rightarrow \dfrac{{2\pi }}{T} = \sqrt {\dfrac{k}{{m{{\left( {\dfrac{b}{a}} \right)}^2}}}}$
On solving further, we get
$T = 2\pi \left( {\dfrac{b}{a}} \right)\sqrt {\dfrac{m}{k}}$
Then we need to match the correct option.
The correct option is C.