$A$ 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 .... $s$.

$A$ 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 total mechanical energy of a spring-mass system in simple harmonic motion is $E = \frac{1}{2}m\omega^2 A^2$. Suppose the oscillating particle is replaced by another particle of double the mass while the amplitude $A$ remains the same. The new mechanical energy will:

$A$ block of mass $5 \ kg$ moves along the $x$-direction subject to the force $F = (-20x + 10) \ N$,with the value of $x$ in metre. At time $t = 0 \ s$,it is at rest at position $x = 1 \ m$. The position and momentum of the block at $t = (\pi / 4) \ s$ are

In the figure,$S_1$ and $S_2$ are identical springs. The oscillation frequency of the mass $m$ is $f$. If one spring is removed,the frequency will become

An assembly of identical spring-mass systems is placed on a smooth horizontal surface as shown. Initially,the springs are relaxed. The left mass is displaced to the left while the right mass is displaced to the right and released. The resulting collision is elastic. The time period of the oscillations of the system is: