Oscillations

${f_A} = 2{f_B}$

$ \Rightarrow {1 \over {2\pi }}\sqrt {{g \over {{l_A}}}} = 2 \times {1 \over {2\pi }}\sqrt {{g \over {{l_B}}}} $

$ \Rightarrow {l \over {{l_A}}} = 4 \times {l \over {{l_B}}}$

$ \Rightarrow {l_A} = {{{l_B}} \over 4}$ , which does not depend on mass.

In $\Delta OAC,\,\cos \theta = A/l$

$ \Rightarrow OA = lcos\theta $

$ \therefore $ $AB = l\left( {1 - \cos \theta } \right) = h$

At point, C the velocity of bob = 0. The vertical acceleration = g

$ \therefore $ ${v^2} = 2gh$

$ \Rightarrow v = \sqrt {2gl\left( {1 - \cos \theta } \right)} $

At equilibrium,

$m g=A l \rho g \Rightarrow m=A \rho l$

where, $l$ length of part immersed in liquid.

When it is given in downward displacement $y$, restoring force (upward direction) on block is

$\begin{aligned} F & =-[A(l+y) \rho g-m g] \\\\ & =-[A(l+y) \rho g-A l \rho g] \\\\ & =-A \rho g y \end{aligned}$

i.e. $F \propto-y$ or $a \propto-\gamma$, so it execute SHM (inertia factor). Mass of block $=m$ Spring factor $=A \rho g$

$\begin{aligned} \text { Time period } & =2 \pi \sqrt{\frac{\text { Inertia factor }}{\text { Spring factor }}} \\\\ T & =2 \pi \sqrt{\frac{m}{A \rho g}} \\\\ \Rightarrow \quad T^2 & \propto \frac{m}{A \rho} \end{aligned}$

Time period of simple pendulum when it is falling under acceleration of ‘a’.

$T=2 \pi \sqrt{\frac{l}{g-a}}$ ..... (i)

In the case of freely falling, $a=g$

$\therefore$ From Eq. (i),

$T=2 \pi \sqrt{\frac{l}{g-g}} \Rightarrow T=\infty$

Given, spring constant $=k$

From law of conservation of energy

$m g h=\frac{1}{2} k x_0^2-m g x_0$

Where, $x_0$ is maximum elongation in spring

$\begin{array}{ll} \Rightarrow & \frac{1}{2} k x_0^2-m g x_0-m g h=0 \\ \Rightarrow & x_0^2-\frac{2 m g}{k} x_0-\frac{2 m g}{k} h=0 \\ & x_0=\frac{\frac{2 m g}{k} \pm \sqrt{\left(\frac{2 m g}{k}\right)^2+4 \times \frac{2 m g}{k} h}}{2} \end{array}$

Amplitude $=$ elongation in spring for lowest extreme position $-$ elongation in spring for equilibrium position

$=x_0-x_1=\frac{m g}{k} \sqrt{1+\frac{2 h k}{m g}}$

$\begin{gathered} \text {} k=\frac{f}{l} \Rightarrow k \propto \frac{1}{l} \\ \Rightarrow \frac{k_1}{k_2}=\frac{l_2}{l_1}=\frac{2}{1} \\ k_1=2 k, k_2=k \end{gathered}$

In series, $\frac{1}{k^{\prime}}=\frac{1}{k_1}+\frac{1}{k_2}=\frac{1}{2 k}+\frac{1}{k}=\frac{3}{2 k}$

$\therefore \quad k^{\prime}=2 / 3 k$