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 in series to parallel combination will be

All the springs in figures $(a)$,$(b)$,and $(c)$ are identical,each having a force constant $K$. $A$ mass $m$ is attached to each system. If $T_a, T_b$,and $T_c$ are the time periods of oscillations of the three systems in figures $(a)$,$(b)$,and $(c)$ respectively,then:

When a block of mass $m$ is suspended separately by two different springs having time periods $t_1$ and $t_2$,respectively. If the same mass is connected to the series combination of both springs,then its time period is given by:

$A$ spring of force constant $k$ is cut into two pieces such that one piece is double the length of the other. Then the long piece will have a force constant of

$A$ coin is placed on a horizontal platform. The platform performs vertical simple harmonic motion with an angular frequency $\omega$. The amplitude of oscillation is gradually increased. At what condition will the coin first lose contact with the platform?

Consider a spring-mass simple harmonic oscillator in one dimension. The mass of the particle is $m \text{ kg}$ and the spring constant is $k \text{ Nm}^{-1}$. At a given instant, the extension of the spring is $x \text{ m}$ and the speed of the particle is $v \text{ ms}^{-1}$. On the $x-v$ plane, if the graph of $v$ as a function of $x$ is a circle, then the correct option is: