$A$ body of mass $m$ is suspended to an ideal spring of force constant $k$. The expected change in the position of the body due to an additional force $F$ acting vertically downwards is

$A$ particle of mass $0.50 \ kg$ executes simple harmonic motion under force $F = -50 \ (N/m) x$. The time period of oscillation is $\frac{x}{35} \ s$. The value of $x$ is . . . . . (Given $\pi = \frac{22}{7}$)

Two masses $m_1 = 1 \, kg$ and $m_2 = 0.5 \, kg$ are suspended together by a massless spring of spring constant $k = 12.5 \, N/m$. When the masses are in equilibrium,$m_1$ is removed without disturbing the system. The new amplitude of oscillation will be .......... $cm$.

When a block of mass $m$ is suspended separately by two different springs having time periods $t_1$ and $t_2$,respectively. If the same mass is connected to the series combination of both springs,then its time period is given by:

$A$ bar of mass $m$ is suspended horizontally on two vertical springs of spring constant $k$ and $3k$. The bar bounces up and down while remaining horizontal. Find the time period of oscillation of the bar (Neglect mass of springs and friction everywhere).

$A$ spring-mass system is oscillating horizontally. What will be the effect on the time period if the spring is made to oscillate vertically?