$A$ particle of mass $200 \,g$ executes $S.H.M.$ The restoring force is provided by a spring of force constant $80 \,N/m$. The time period of oscillations is .... $s$.

$A$ particle at the end of a spring executes simple harmonic motion with a period $t_1$,while the corresponding period for another spring is $t_2$. If the period of oscillation with the two springs in series is $T$,then

$A$ weightless spring of length $60\, cm$ and force constant $200\, N/m$ is kept straight and unstretched on a smooth horizontal table and its ends are rigidly fixed. $A$ mass of $0.25\, kg$ is attached at the middle of the spring and is slightly displaced along the length. The time period of the oscillation of the mass is

The force constant of a weightless spring is $16 \, N/m$. $A$ body of mass $1.0 \, kg$ suspended from it is pulled down through $5 \, cm$ and then released. The maximum kinetic energy of the system (spring + body) will be

Frequency of a particle performing $S.H.M.$ is $10 \ Hz$. The particle is suspended from a vertical spring. At the highest point of its oscillation,the spring is unstretched. The maximum speed of the particle is $(g = 10 \ m/s^2)$.

$A$ mass $x \ g$ is suspended from a light spring. It is pulled in a downward direction and released so that the mass performs $S.H.M.$ of period $T$. If the mass is increased by $Y \ g$,the period becomes $4T/3$. The ratio of $Y/x$ is: