$A$ mass $m$ performs oscillations of period $T$ when suspended by a spring of force constant $K$. If the spring is cut into two equal parts and arranged in parallel,and the same mass $m$ is oscillated by them,then the new time 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:

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 ...... $s$.

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 '$m$' attached to a spring oscillates with a period of $3 \ s$. If the mass is increased by $0.6 \ kg$,the period increases by $3 \ s$. The initial mass '$m$' is equal to (in $kg$)

$A$ body of mass $5\; kg$ hangs from a spring and oscillates with a time period of $2\pi\; s$. If the body is removed,the length of the spring will decrease by: