$A$ $10 \,kg$ metal block is attached to a spring of spring constant $1000 \,N \,m^{-1}$. The block is displaced from the equilibrium position by $10 \,cm$ and released. The maximum acceleration of the block is: (in $\,m/s^2$)

The frequency of oscillation of a mass $m$ suspended by a spring is $v_1$. If the length of the spring is cut to half, the same mass oscillates with frequency $v_2$. The value of $v_2/v_1$ 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 $m$ is suspended separately by two different springs having time periods $t_1$ and $t_2$. If the same mass is connected to a parallel combination of both springs,then its time period will be

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

$A$ body of mass $m$ is attached to one end of a massless spring which is suspended vertically from a fixed point. The mass is held in hand,so that the spring is neither stretched nor compressed. Suddenly,the support of the hand is removed. The lowest position attained by the mass during oscillation is $4 \, cm$ below the point where it was held in hand.
$(a)$ What is the amplitude of oscillation?
$(b)$ Find the frequency of oscillation.