To make the frequency of an oscillator double,we have to

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

$A$ light spring is suspended with mass $m_1$ at its lower end and its upper end fixed to a rigid support. The mass is pulled down a short distance and then released. The period of oscillation is $T$ seconds. When a mass $m_2$ is added to $m_1$ and the system is made to oscillate,the period is found to be $\frac{3}{2} T$. The ratio $m_1 : m_2$ is

On a smooth inclined plane,a block of mass $M$ is fixed to two rigid supports using two springs,as shown in the figure. If each spring has spring constant $k$,then the period of oscillation of the block is (Neglect the masses of the springs)

The springs in the figure are identical,but the length of the spring in $A$ is three times that in $B$. The ratio of the time periods $T_A/T_B$ is:

$A$ mass $m_1$ performs $S.H.M.$ with amplitude $A$ while connected to a horizontal spring. While mass $m_1$ is passing through the mean position, another mass $m_2$ $(m_2 < m_1)$ is placed on it so that both masses move together with amplitude $A_1$. The ratio of $A/A_1$ is