$A$ mass $m$ is suspended from a spring of length $l$ and force constant $K$. The frequency of vibration of the mass is $f_1$. The spring is cut into two equal parts and the same mass is suspended from one of the parts. The new frequency of vibration of the mass is $f_2$. Which of the following relations between the frequencies is correct?

$A$ small mass $m$, attached to one end of a spring with a negligible mass and an unstretched length $L$, executes vertical oscillations with angular frequency $\omega_{0}$. When the mass is rotated with an angular speed $\omega$ by holding the other end of the spring at a fixed point, the mass moves uniformly in a circular path in a horizontal plane. Then the increase in length of the spring during the rotation is

$A$ body executes $SHM$ under the action of force $F_1$ with time period $T_1$. If the force is changed to $F_2$,it executes $SHM$ with time period $T_2$. If both the forces $F_1$ and $F_2$ act simultaneously in the same direction on the body,its time period is:

$A$ $1 \, kg$ block attached to a spring vibrates with a frequency of $1 \, Hz$ on a frictionless horizontal table. Two springs identical to the original spring are attached in parallel to an $8 \, kg$ block placed on the same table. The frequency of vibration of the $8 \, kg$ block is ..... $Hz$.

An object is attached to the bottom of a light vertical spring and set vibrating. The maximum speed of the object is $15 \,cm/s$ and the period is $628 \,ms$. The amplitude of the motion in $cm$ is:

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 .......... $cm$.