$A$ 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

$A$ block of mass $M$ hangs from a spring and oscillates vertically with an angular frequency $\omega$. If the block is removed from the spring,when it is in equilibrium position,the spring shortens by

$A$ mass $m$ attached to a spring oscillates with a period of $3\,s$. If the mass is increased by $1\,kg$,the period increases by $1\,s$. The initial mass $m$ is

The scale of a spring balance which can measure from $0$ to $15 \ kg$ is $0.25 \ m$ long. If a body suspended from this balance oscillates with a time period $\frac{2 \pi}{5} \ s$,neglecting the mass of the spring,find the mass of the body suspended. (in $kg$)

Figure $(a)$ shows a spring of force constant $k$ clamped rigidly at one end and a mass $m$ attached to its free end. $A$ force $F$ applied at the free end stretches the spring. Figure $(b)$ shows the same spring with both ends free and attached to a mass $m$ at either end. Each end of the spring in Figure $(b)$ is stretched by the same force $F$.
$(a)$ What is the maximum extension of the spring in the two cases?
$(b)$ If the mass in Figure $(a)$ and the two masses in Figure $(b)$ are released,what is the period of oscillation in each case?

Consider two identical springs each of spring constant $k$ and negligible mass compared to the mass $M$ as shown. Fig. $1$ shows one of them and Fig. $2$ shows their series combination. The ratio of the time period of oscillation of the two $SHM$ is $\frac{T_b}{T_a} = \sqrt{x}$,where the value of $x$ is (Round off to the Nearest Integer).