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:

Two masses $m_1$ and $m_2$ are suspended together by a massless spring of constant $k$. When the masses are in equilibrium,$m_1$ is removed without disturbing the system. Then the angular frequency of oscillation of $m_2$ is

$A$ mass $m = 1.0\,kg$ is placed on a flat pan attached to a vertical spring fixed on the ground. The mass of the spring and the pan is negligible. When pressed slightly and released,the mass executes simple harmonic motion. The spring constant is $k = 500\,N/m$. What is the amplitude $A$ of the motion,so that the mass $m$ tends to get detached from the pan? (Take $g = 10\,m/s^2$).

How does the period of oscillation depend on the mass of the block attached to the end of a spring?

$A$ mass at the end of a spring executes harmonic motion about an equilibrium position with an amplitude $A$. Its speed as it passes through the equilibrium position is $V$. If extended $2A$ and released,the speed of the mass passing through the equilibrium position will be

$A$ mass $m$ is suspended separately by two different springs of spring constant $K_1$ and $K_2$ giving the time periods $t_1$ and $t_2$ respectively. If the same mass $m$ is connected by both springs as shown in the figure,then the time period $t$ is given by the relation: