The potential energy of a particle of mass $m$ situated in a unidimensional potential field varies as $U(x) = U_0(1 - \cos ax)$,where $U_0$ and $a$ are constants. The time period of small oscillations of the particle about the mean position is

Consider a spring-mass simple harmonic oscillator in one dimension. The mass of the particle is $m \text{ kg}$ and the spring constant is $k \text{ Nm}^{-1}$. At a given instant, the extension of the spring is $x \text{ m}$ and the speed of the particle is $v \text{ ms}^{-1}$. On the $x-v$ plane, if the graph of $v$ as a function of $x$ is a circle, then the correct option is:

$A$ spring has a certain mass suspended from it and its period for vertical oscillations is $T_1$. The spring is now cut into two equal halves and the same mass is suspended from one of the halves. The period of vertical oscillations is now $T_2$. The ratio $T_1 / T_2$ is

$A$ $100 \,g$ mass stretches a particular spring by $9.8 \,cm$,when suspended vertically from it. How large a mass must be attached to the spring if the period of vibration is to be $6.28 \,s$?

When a mass $m$ is attached to a spring,it extends by $0.2 \ m$. The mass $m$ is given a slight additional extension and released. Find the time period (take $g = \pi^2 \ m/s^2$).

On a frictionless horizontal plane,a bob of mass $m=0.1 \text{ kg}$ is attached to a spring with natural length $l_0=0.1 \text{ m}$. The spring constant is $k_1=0.009 \text{ N/m}$ when the length of the spring $l > l_0$ and is $k_2=0.016 \text{ N/m}$ when $l < l_0$. Initially,the bob is released from $l=0.15 \text{ m}$. Assume that Hooke's law remains valid throughout the motion. If the time period of the full oscillation is $T=(n \pi) \text{ s}$,then the integer closest to $n$ is: