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) = y₀ sin² $$\omega $$t, where 'y' is measured from the lower end of unstretched spring. Then $$\omega $$ is:

A circuit to verify Ohm's law uses ammeter and voltmeter in series or parallel connected correctly to the resistor. In the circuit :

Two identical electric point dipoles have dipole moments $${\overrightarrow p _1} = p\widehat i$$ and $${\overrightarrow p _2} = - p\widehat i$$ and are held on the x axis at distance '$$a$$' from each other. When released, they move along the x-axis with the direction of their dipole moments remaining unchanged. If the mass of each dipole is 'm', their speed when they are infinitely far apart is :

Particle A of mass m₁ moving with velocity $$\left( {\sqrt3\widehat i + \widehat j} \right)m{s^{ - 1}}$$ collides with another particle B of mass m₂ which is at rest initially. Let $$\overrightarrow {{V_1}} $$ and $$\overrightarrow {{V_2}} $$ be the velocities of particles A and B after collision respectively. If m₁ = 2m₂ and after
collision $$\overrightarrow {{V_1}} = $$$$\left( {\widehat i + \sqrt 3 \widehat j} \right)$$ , the angle between $$\overrightarrow {{V_1}} $$ and $$\overrightarrow {{V_2}} $$ is :