$A$ clock $S$ is based on the oscillations of a spring,and a clock $P$ is based on pendulum motion. Both clocks run at the same rate on Earth. On a planet having the same density as Earth but twice the radius,then:

Two identical balls $A$ and $B$,each of mass $0.1 \ kg$,are attached to two identical massless springs. The spring-mass system is constrained to move inside a rigid smooth pipe bent in the form of a circle as shown in the figure. The pipe is fixed in a horizontal plane. The centers of the balls can move in a circle of radius $0.06 \ m$. Each spring has a natural length of $0.06\pi \ m$ and a force constant of $0.1 \ N/m$. Initially,both balls are displaced by an angle $\theta = \pi/6$ radian with respect to the diameter $PQ$ of the circle and released from rest. The frequency of oscillation of the ball $B$ is:

$A$ block of mass $(10 \alpha) \text{ g}$,where $\alpha$ is a constant,is moving with velocity $3 \text{ m/s}$ to the right. It collides inelastically with a block on the right of mass $10 \text{ g}$ and sticks to it. The right block is connected to three springs as shown in the figure. The spring constant of each spring is $k = 2 \text{ N/m}$. If the amplitude of the resulting simple harmonic motion is $A = \frac{1}{2\sqrt{2}} \text{ m}$,then the value of $\alpha$ is:

Two particles,$1$ and $2$,each of mass $m$,are connected by a massless spring and are on a horizontal frictionless plane,as shown in the figure. Initially,the two particles,with their center of mass at $x_0$,are oscillating with amplitude $a$ and angular frequency $\omega$. Thus,their positions at time $t$ are given by $x_1(t) = (x_0 + d) + a \sin \omega t$ and $x_2(t) = (x_0 - d) - a \sin \omega t$,respectively,where $d > 2a$. Particle $3$ of mass $m$ moves towards this system with speed $u_0 = a \omega / 2$ and undergoes an instantaneous elastic collision with particle $2$ at time $t_0$. Finally,particles $1$ and $2$ acquire a center of mass speed $v_{cm}$ and oscillate with amplitude $b$ and the same angular frequency.
$(1)$ If the collision occurs at time $t_0 = 0$,the value of $v_{cm} / (a \omega)$ will be
$(2)$ If the collision occurs at time $t_0 = \pi / (2 \omega)$,then the value of $4b^2 / a^2$ will be

In the figure shown,there is friction between the blocks $P$ and $Q$,but the contact between the block $Q$ and the lower surface is frictionless. Initially,the block $Q$ with block $P$ over it lies at $x=0$,with the spring at its natural length. The block $Q$ is pulled to the right and then released. As the spring-block system undergoes $S.H.M.$ with amplitude $A$,the block $P$ tends to slip over $Q$. $P$ is more likely to slip at:

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