Two identical springs of constant $K$ are connected in series and parallel as shown in the figure. $A$ mass $m$ is suspended from them. The ratio of their frequencies of vertical oscillations will be

$A$ small mass $m$, attached to one end of a spring with a negligible mass and an unstretched length $L$, executes vertical oscillations with angular frequency $\omega_{0}$. When the mass is rotated with an angular speed $\omega$ by holding the other end of the spring at a fixed point, the mass moves uniformly in a circular path in a horizontal plane. Then the increase in length of the spring during the rotation is

The time period of simple harmonic motion of mass $M$ in the given figure is $\pi \sqrt{\frac{\alpha M}{5 K}}$,where the value of $\alpha$ is . . . . . . .

$A$ mass of $0.2\,kg$ is attached to the lower end of a massless spring of force-constant $200\,N/m,$ the upper end of which is fixed to a rigid support. Which of the following statements is/are true?

The total mechanical energy of a spring-mass system in simple harmonic motion is $E = \frac{1}{2}m\omega^2 A^2$. Suppose the oscillating particle is replaced by another particle of double the mass while the amplitude $A$ remains the same. The new mechanical energy will:

$A$ system consisting of three masses attached to a spring is in equilibrium. When the $700\,g$ mass is removed,the system oscillates with a period of $3\,s$. When the $500\,g$ mass is also removed,what will be the new time period of the system? (in $\text{seconds}$)