To make the frequency of a spring oscillator double,we have to

The motion of a mass on a spring,with spring constant $K$ is as shown in the figure. The equation of motion is given by $x(t) = A \sin \omega t + B \cos \omega t$ with $\omega = \sqrt{\frac{K}{m}}$. Suppose that at time $t = 0$,the position of the mass is $x(0)$ and velocity is $v(0)$,then its displacement can also be represented as $x(t) = C \cos (\omega t - \phi)$,where $C$ and $\phi$ are:

The period of oscillation of a mass $M$ suspended from a spring of negligible mass is $T$. If along with it another mass $M$ is also suspended,the period of oscillation will now be

Two bodies $A$ and $B$ of equal masses are suspended from two separate springs of force constants $k_1$ and $k_2$ respectively. If the two bodies oscillate such that their maximum velocities are equal,the ratio of the amplitudes of oscillation of $A$ and $B$ will be

Two identical springs are connected to a mass $m$ as shown in the figure ($k$ = spring constant). If the time period of the configuration in $(a)$ is $2 \,s$, what is the time period of the configuration in $(b)$?

Figure $(a)$ shows a spring of force constant $k$ clamped rigidly at one end and a mass $m$ attached to its free end. $A$ force $F$ applied at the free end stretches the spring. Figure $(b)$ shows the same spring with both ends free and attached to a mass $m$ at either end. Each end of the spring in Figure $(b)$ is stretched by the same force $F$.
$(a)$ What is the maximum extension of the spring in the two cases?
$(b)$ If the mass in Figure $(a)$ and the two masses in Figure $(b)$ are released,what is the period of oscillation in each case?