Two massless springs with spring constants $2k$ and $9k$ carry $50\,g$ and $100\,g$ masses at their free ends. These two masses oscillate vertically such that their maximum velocities are equal. Then,the ratio of their respective amplitudes will be

The vertical extension in a light spring by a weight of $1\, kg$ suspended from it is $9.8\, cm$. What is the period of oscillation?

$A$ 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

In figure $(A)$,mass '$2m$' is fixed on mass '$m$' which is attached to two springs of spring constant $k$. In figure $(B)$,mass '$m$' is attached to two springs of spring constant '$k$' and '$2k$'. If mass '$m$' in $(A)$ and $(B)$ are displaced by distance '$x$' horizontally and then released,then the time periods $T_{1}$ and $T_{2}$ corresponding to $(A)$ and $(B)$ respectively follow the relation.

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).

$A$ simple harmonic oscillator consists of a particle of mass $m$ and an ideal spring with spring constant $k$. The particle oscillates with a time period $T$. The spring is cut into two equal parts. If one part oscillates with the same particle, the new time period will be