$A$ $100 \,g$ mass stretches a particular spring by $9.8 \,cm$,when suspended vertically from it. How large a mass must be attached to the spring if the period of vibration is to be $6.28 \,s$?

$A$ spring of force constant $k$ is cut into two pieces such that one piece is double the length of the other. Then the long piece will have a force constant of

All the springs in figures $(a)$,$(b)$,and $(c)$ are identical,each having a force constant $K$. $A$ mass $m$ is attached to each system. If $T_a, T_b$,and $T_c$ are the time periods of oscillations of the three systems in figures $(a)$,$(b)$,and $(c)$ respectively,then:

$A$ spring has a certain mass suspended from it and its period for vertical oscillations is $T_1$. The spring is now cut into two equal halves and the same mass is suspended from one of the halves. The period of vertical oscillations is now $T_2$. The ratio $T_1 / T_2$ is

Inside a lift, a spring (force constant $k = 1000 \ N/m$) and a block $(mass = 1 \ kg)$ are both in a state of rest. Now, the lift suddenly starts moving upwards with an acceleration $a = g$. Find the maximum total compression in the spring in centimeters. $(g = 10 \ m/s^2)$

The restoring force of a spring with a block attached to the free end of the spring is represented by