$A$ force of $6.4 \,N$ stretches a vertical spring by $0.1 \,m$. If it were to oscillate with a period of $\frac{\pi}{4} \,s$, then the mass that is to be suspended from the spring is:

For a simple harmonic motion in a mass-spring system shown,the surface is frictionless. When the mass of the block is $1\,kg$,the angular frequency is $\omega_1$. When the mass of the block is $2\,kg$,the angular frequency is $\omega_2$. The ratio $\omega_2 / \omega_1$ is

$A$ 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 ..... .

$A$ body of mass $64 \ g$ is made to oscillate turn by turn on two different springs $A$ and $B$. Spring $A$ and $B$ have force constants $4 \ N/m$ and $16 \ N/m$ respectively. If $T_{1}$ and $T_{2}$ are the periods of oscillation of springs $A$ and $B$ respectively,then $\frac{T_{1}+T_{2}}{T_{1}-T_{2}}$ will be:

$A$ 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:

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