The frequency of oscillation of a mass $m$ suspended by a spring is $v$. If the mass is reduced to one-fourth,what will be the new frequency of oscillation?

When a particle of mass $m$ is attached to a vertical spring of spring constant $k$ and released,its motion is described by $y(t) = y_{0} \sin^{2} \omega t$,where $y$ is measured from the lower end of the unstretched spring. Then $\omega$ is

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

Two bodies $A$ and $B$ of equal mass are suspended from two massless springs of spring constant $k_1$ and $k_2$,respectively. If the bodies oscillate vertically such that their amplitudes are equal,the ratio of the maximum velocity of $A$ to the maximum velocity of $B$ is

$A$ mass $m$ is suspended from two coupled springs connected in series. The force constants for the springs are $K_1$ and $K_2$. The time period of the suspended mass will be:

$A$ particle of mass $m$ is attached to one end of a massless spring of force constant $k$,lying on a frictionless horizontal plane. The other end of the spring is fixed. The particle starts moving horizontally from its equilibrium position at time $t=0$ with an initial velocity $u_0$. When the speed of the particle is $0.5 u_0$,it collides elastically with a rigid wall. After this collision:
$(A)$ the speed of the particle when it returns to its equilibrium position is $u_0$.
$(B)$ the time at which the particle passes through the equilibrium position for the first time is $t=\pi \sqrt{\frac{m}{k}}$.
$(C)$ the time at which the maximum compression of the spring occurs is $t =\frac{4 \pi}{3} \sqrt{\frac{m}{k}}$.
$(D)$ the time at which the particle passes through the equilibrium position for the second time is $t=\frac{5 \pi}{3} \sqrt{\frac{m}{k}}$.