$A$ mass $M$ is suspended by two springs of force constants $K_1$ and $K_2$ respectively,as shown in the diagram. The total elongation (stretch) of the two springs is

$A$ block of mass $m$ attached to a massless spring is performing oscillatory motion of amplitude $A$ on a frictionless horizontal plane. If half of the mass of the block breaks off when it is passing through its equilibrium point,the amplitude of oscillation for the remaining system becomes $fA$. The value of $f$ is

$A$ block of mass $2\,kg$ is attached to two identical springs,each with a spring constant of $20\,N/m$. The block is placed on a frictionless surface,and the outer ends of the springs are attached to rigid supports (see figure). When the mass is displaced from its equilibrium position,it executes simple harmonic motion. The time period of oscillation is $\frac{\pi}{\sqrt{x}}$ in $SI$ units. The value of $x$ is $..........$

As shown in the figure, a spring with spring constant $k = 150 \ N/m$ is kept in a stretched position by holding two masses of $1 \ kg$ and $200 \ g$ at a separation greater than the natural length of the spring. When the masses are released, assuming the horizontal surface to be frictionless, the angular frequency (in $SI$ units) of the system is:

$A$ mass $M$,attached to a horizontal spring,executes $S.H.M.$ with amplitude $A_1$. When the mass $M$ passes through its mean position,a smaller mass $m$ is placed over it and both of them move together with amplitude $A_2$. The ratio of $\frac{A_1}{A_2}$ is

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: