The frequency of oscillation of a mass $m$ suspended by a spring is $v_1$. If the length of the spring is cut to one-third,then the same mass oscillates with frequency $v_2$. Then:

$A$ mass $M$ is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes $S.H.M.$ of period $T$. If the mass is increased by $m$,the time period becomes $\frac{5T}{3}$. What is the ratio $\left(\frac{M}{m}\right)$?

$A$ mass $m$ is suspended separately by two springs of spring constants $k_1$ and $k_2$. The time periods of oscillations in the two cases are $T_1$ and $T_2$ respectively. If the same mass $m$ is suspended by connecting the two springs in parallel (as shown in the figure),then the time period of the oscillation is $T$. The correct relation is:

$A$ load of mass $m$ falls from a height $h$ onto a scale pan hung from a spring as shown in the figure. If the spring constant is $k$,the mass of the scale pan is zero,and the mass $m$ does not bounce relative to the pan,then the amplitude of vibration is

Consider two identical springs each of spring constant $k$ and negligible mass compared to the mass $M$ as shown. Fig. $1$ shows one of them and Fig. $2$ shows their series combination. The ratio of the time period of oscillation of the two $SHM$ is $\frac{T_b}{T_a} = \sqrt{x}$,where the value of $x$ is (Round off to the Nearest Integer).

$A$ block of mass $m$ has two similar rubber ribbons attached to it as shown in the figure. The force constant of each rubber ribbon is $K$ and the surface is frictionless. The block is displaced from the mean position by $x$ and released. At the mean position,the ribbons are undeformed. The vibration period is