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
$f$
$f \times 2$
$f \times \sqrt 2 $
$f/\sqrt 2 $
If two similar springs each of spring constant $K _{1}$ are joined in series, the new spring constant and time period would be changed by a factor
The length of a spring is $l$ and its force constant is $k$. When a weight $W$ is suspended from it, its length increases by $x$. If the spring is cut into two equal parts and put in parallel and the same weight $W$ is suspended from them, then the extension will be
A mass of $2.0\, kg$ is put on a flat pan attached to a vertical spring fixed on the ground as shown in the figure. The mass of the spring and the pan is negligible. When pressed slightly and released the mass executes a simple harmonic motion. The spring constant is $200\, N/m.$ What should be the minimum amplitude of the motion so that the mass gets detached from the pan (take $g = 10 m/s^2$).
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 :-
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 ratios of time period of oscillation of the two $SHM$ is $\frac{ T _{ b }}{ T _{ a }}=\sqrt{ x },$ where value of $x$ is
(Round off to the Nearest Integer)