$A$ 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, $g = 10 \,m/s^2$)

Two oscillating systems,a simple pendulum and a vertical spring-mass system,have the same time period of motion on the surface of the Earth. If both are taken to the moon,then-

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

$A$ particle of mass $m$ moves in the potential energy $U(x)$ shown in the figure. The potential energy is given by $U = \frac{1}{2}kx^2$ for $x < 0$ and $U = mgx$ for $x > 0$. The period of the motion when the particle has total energy $E$ is

$A$ man weighing $60\ kg$ stands on the horizontal platform of a spring balance. The platform starts executing simple harmonic motion of amplitude $0.1\ m$ and frequency $\frac{2}{\pi}\ Hz$. Which of the following statements is correct?

$A$ 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: