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

A mass $m$ is suspended from the two coupled springs connected in series. The force constant for springs are ${K_1}$ and ${K_2}$. The time period of the suspended mass will be

Find the time period of mass $M$ when displaced from its equilibrium position and  then released for the system shown in figure.

Two identical springs have the same force constant $73.5 \,Nm ^{-1}$. The elongation produced in each spring in three cases shown in Figure-$1$, Figure-$2$ and Figure-$3$ are $\left(g=9.8 \,ms ^{-2}\right)$

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