Due to some force $F_1$,a body oscillates with a period of $4/5 \, s$,and due to another force $F_2$,it oscillates with a period of $3/5 \, s$. If both forces act simultaneously,the new period will be .... $s$.

Match the following physical quantities for a particle executing Simple Harmonic Motion $(SHM)$ given by $y = A \sin(\omega t)$:
$(a)$ Velocity $(v)$
$(b)$ Potential Energy $(PE)$
$(c)$ Total Energy $(TE)$
$(d)$ Acceleration $(a)$
$(i)$ Constant
(ii) $A\omega \cos(\omega t)$
(iii) $\frac{1}{2} k A^2 \sin^2(\omega t)$
(iv) $-\omega^2 y$

$A$ system is oscillating with undamped simple harmonic motion. Then the

$A$ vertical spring-mass system has the same time period as a simple pendulum undergoing small oscillations. Now, both of them are placed in an elevator moving downwards with an acceleration $a = 5 \,m/s^2$. The ratio of the time period of the spring-mass system to the time period of the pendulum is (Assume, acceleration due to gravity, $g = 10 \,m/s^2$)

An oscillator of mass $M$ is at rest in its equilibrium position in a potential $V = \frac{1}{2}k(x - X)^2$. $A$ particle of mass $m$ comes from the right with speed $u$ and collides completely inelastically with $M$ and sticks to it. This process repeats every time the oscillator crosses its equilibrium position. The amplitude of oscillations after $13$ collisions is: $(M = 10, m = 5, u = 1, k = 1)$.

Consider two identical cylinders (each of mass $m$,density $\rho_0$,horizontal cross-section area $s$) in equilibrium,partially submerged in two containers filled with liquids of densities $\rho_1$ and $\rho_2$ as shown in the figure. Find the period of small oscillations of this system about its equilibrium. Neglect the changes in the level of liquids in the containers. Neglect the mass of the strings. The acceleration due to gravity is $g$. ($v$ is the volume of each block).