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

$A$ mass $m$ is suspended separately by two different springs,and the time periods are $t_1$ and $t_2$ respectively. If it is connected by both springs in parallel as shown in the figure,then the time period is $t_0$. The correct relation is:

$A$ mass $m$ is suspended from a spring of force constant $k$ and just touches another identical spring fixed to the floor as shown in the figure. The time period of small oscillations is

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 object of mass $m$ is performing simple harmonic motion on a smooth horizontal surface as shown in the figure. Just as the oscillating object reaches its extreme position,another object of mass $2m$ is placed gently on the oscillating object,and it sticks to it. Consider the following statements:
$(a)$ Amplitude of oscillation remains unchanged.
$(b)$ Time period of oscillation remains unchanged.
$(c)$ Total mechanical energy of the system does not change.
$(d)$ The maximum speed of the oscillating object changes.
Which of the above statements are correct?

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: