Write the formula for the time period $(T)$ of a particle executing Simple Harmonic Motion $(SHM)$.

The equivalent spring constant of the system as shown in the figure will be:

Two springs of force constants $300 \, N/m$ (Spring $A$) and $400 \, N/m$ (Spring $B$) are joined together in series. The combination is compressed by $8.75 \, cm$. The ratio of energy stored in $A$ and $B$ is $\frac{E_A}{E_B}$. Then $\frac{E_A}{E_B}$ is equal to

All the springs in figures $(a)$,$(b)$,and $(c)$ are identical,each having a force constant $K$. $A$ mass $m$ is attached to each system. If $T_a, T_b$,and $T_c$ are the time periods of oscillations of the three systems in figures $(a)$,$(b)$,and $(c)$ respectively,then:

$A$ particle of mass $m$ is attached to one end of a massless spring of force constant $k$,lying on a frictionless horizontal plane. The other end of the spring is fixed. The particle starts moving horizontally from its equilibrium position at time $t=0$ with an initial velocity $u_0$. When the speed of the particle is $0.5 u_0$,it collides elastically with a rigid wall. After this collision:
$(A)$ the speed of the particle when it returns to its equilibrium position is $u_0$.
$(B)$ the time at which the particle passes through the equilibrium position for the first time is $t=\pi \sqrt{\frac{m}{k}}$.
$(C)$ the time at which the maximum compression of the spring occurs is $t =\frac{4 \pi}{3} \sqrt{\frac{m}{k}}$.
$(D)$ the time at which the particle passes through the equilibrium position for the second time is $t=\frac{5 \pi}{3} \sqrt{\frac{m}{k}}$.

$A$ 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