$A$ particle of mass $1 \, \text{kg}$ is hanging from a spring of force constant $100 \, \text{N/m}$. The mass is pulled slightly downward and released so that it executes free simple harmonic motion with time period $T$. The time when the kinetic energy and potential energy of the system will become equal is $\frac{T}{x}$. The value of $x$ is ..... .

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 $, s$)?

On a frictionless horizontal plane,a bob of mass $m=0.1 \text{ kg}$ is attached to a spring with natural length $l_0=0.1 \text{ m}$. The spring constant is $k_1=0.009 \text{ N/m}$ when the length of the spring $l > l_0$ and is $k_2=0.016 \text{ N/m}$ when $l < l_0$. Initially,the bob is released from $l=0.15 \text{ m}$. Assume that Hooke's law remains valid throughout the motion. If the time period of the full oscillation is $T=(n \pi) \text{ s}$,then the integer closest to $n$ is:

$A$ block of mass $2\,kg$ is attached to two identical springs,each with a spring constant of $20\,N/m$. The block is placed on a frictionless surface,and the outer ends of the springs are attached to rigid supports (see figure). When the mass is displaced from its equilibrium position,it executes simple harmonic motion. The time period of oscillation is $\frac{\pi}{\sqrt{x}}$ in $SI$ units. The value of $x$ is $..........$

$A$ 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:

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