The time period of simple harmonic motion of mass $M$ in the given figure is $\pi \sqrt{\frac{\alpha M}{5 K}}$,where the value of $\alpha$ is . . . . . . .

$A$ 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 $t=\frac{5 \pi}{3} \sqrt{\frac{m}{k}}$.

When a body of mass $8 \,kg$ is attached to a spring balance, the reading of the balance is $20 \,cm$. Instead of $8 \,kg$, if another body of mass $M$ is suspended from the spring balance and is made to oscillate vertically, the time period of oscillation is $\frac{\pi}{5} \,s$, then the value of $M$ is (Acceleration due to gravity $= 10 \,m/s^2$) (in $\,kg$)

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 $m$ is attached to two springs as shown in the figure. The spring constants of the two springs are $K_1$ and $K_2$. For the frictionless surface,the time period of oscillation of mass $m$ is

In the following questions,match Column-$I$ with Column-$II$ and choose the correct options.