When a body of mass $m$ is suspended from a spiral spring,the spring stretches by a distance of $20\, cm$. If the spring is further stretched by a small distance and released,what is the period of oscillation (in $, s$)?

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 of vertical oscillations will be

In figure $(A)$,mass '$2m$' is fixed on mass '$m$' which is attached to two springs of spring constant $k$. In figure $(B)$,mass '$m$' is attached to two springs of spring constant '$k$' and '$2k$'. If mass '$m$' in $(A)$ and $(B)$ are displaced by distance '$x$' horizontally and then released,then the time periods $T_{1}$ and $T_{2}$ corresponding to $(A)$ and $(B)$ respectively follow the relation.

$A$ 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

Two springs,of force constants $k_1$ and $k_2$,are connected to a mass $m$ as shown. The frequency of oscillation of the mass is $f$. If both $k_1$ and $k_2$ are made four times their original values,the frequency of oscillation becomes

Let $T_1$ and $T_2$ be the time periods of two springs $A$ and $B$ when a mass $m$ is suspended from them separately. Now both the springs are connected in parallel and the same mass $m$ is suspended with them. If $T$ is the new time period in this position,then: