Two identical springs of constant $k$ are connected in series and then in parallel. $A$ mass $m$ is suspended from them,the ratio of their frequencies of vertical oscillations will be

When a mass $m$ is attached to a spring,it normally extends by $0.2\, m$. If the mass $m$ is given a slight additional extension and released,what will be its time period?

Three masses $700 \, g, 500 \, g$,and $400 \, g$ are suspended at the end of a spring as shown and are 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,it will oscillate with a period of ...... $s$.

Two massless springs of force constants $K_1$ and $K_2$ are joined end to end. The resultant force constant $K$ of the system is

$A$ mass $m$ is suspended by means of two coiled springs which have the same length in the unstretched condition,as shown in the figure. Their force constants are $k_1$ and $k_2$ respectively. When set into vertical vibrations,the period will be:

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: