$A$ spring of length $l$ and force constant $k$ is attached to a mass $m$ and oscillates with frequency $f_1$. If the spring is cut into two equal parts and one part is attached to the same mass $m$ to oscillate,its frequency becomes $f_2$. Find the relation between $f_1$ and $f_2$.

$A$ body of mass $4.9 \, kg$ hangs from a spring and oscillates with a period $0.5 \, s$. On the removal of the body, the spring is shortened by (take $g=10 \, m/s^2, \pi^2=10$)

The scale of a spring balance reading from $0$ to $10 \, kg$ is $0.25 \, m$ long. $A$ body suspended from the balance oscillates vertically with a period of $\pi / 10 \, s$. The mass suspended is ..... $kg$ (neglect the mass of the spring).

$A$ particle connected to the end of a spring executes $S$.$H$.$M$. with period $T_1$. While the corresponding period for another spring is $T_2$. If the period of oscillation with two springs in series is $T$,then

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:

$A$ mass attached to a spring is free to oscillate,with angular velocity $\omega$,in a horizontal plane without friction or damping. It is pulled to a distance $x_{0}$ and pushed towards the centre with a velocity $v_{0}$ at time $t=0$. Determine the amplitude of the resulting oscillations in terms of the parameters $\omega, x_{0}$ and $v_{0}$. [Hint: Start with the equation $x=A \cos (\omega t+\theta)$ and note that the initial velocity is negative.]