$A$ mass '$m_1$' is suspended from a spring of negligible mass. The spring is pulled slightly in the downward direction and released; the mass performs $S.H.M.$ with a period '$T_1$'. If the mass is increased by '$m_2$',the time period becomes '$T_2$'. The ratio $\frac{m_2}{m_1}$ is

The motion of a mass on a spring,with spring constant $K$ is as shown in the figure. The equation of motion is given by $x(t) = A \sin \omega t + B \cos \omega t$ with $\omega = \sqrt{\frac{K}{m}}$. Suppose that at time $t = 0$,the position of the mass is $x(0)$ and velocity is $v(0)$,then its displacement can also be represented as $x(t) = C \cos (\omega t - \phi)$,where $C$ and $\phi$ are:

If two similar springs each of spring constant $K$ are joined in series,the new spring constant and time period would be changed by a factor of:

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:

$A$ mass on a vertical spring begins its motion at rest at $y = 0 \ cm$. It reaches a maximum height of $y = 10 \ cm$. The two forces acting on the mass are gravity and the spring force. The graph of its kinetic energy $(KE)$ versus position is given below. Net force on the mass varies with $y$ as

$A$ stone of mass $m$ is tied to an elastic string of negligible mass and spring constant $k$. The unstretched length of the string is $L$. The other end of the string is fixed to a nail at a point $P$. Initially,the stone is held at the same level as point $P$ and is dropped vertically.
$(a)$ Find the distance $y$ from the top when the mass comes to rest for an instant,for the first time.
$(b)$ What is the maximum velocity attained by the stone in this drop?
$(c)$ What shall be the nature of the motion after the stone has reached its lowest point?