A particle executing linear SHM has velocities $v_1$ and $v_2$ at distances $x_1$ and $x_2$ , respectively, from the mean position. The angular velocity of the particle is

$\sqrt{\frac{v_2^2-v_1^2}{x_1^2 - x_2^2}}$

$\sqrt{\frac{x_1^2 - x_2^2}{v_2^2-v_1^2}} $

$\sqrt{\frac{v_2^2-v_1^2}{x_1^2 - x_2^2}}$

$\sqrt{\frac{x_1^2 + x_2^2}{v_2^2 + v_1^2}} $

$\sqrt{\frac{v_2^2 + v_1^2}{x_1^2 + x_2^2}}$

In simple harmonic motion, the acceleration of a particle is zero when its

both velocity and displacement are zero

both velocity and displacement are maximum

$\sqrt{\frac{v_2^2-v_1^2}{x_1^2 - x_2^2}}$

A spring-and-block oscillator with an ideal spring of force constant 180 N/m oscillates with a frequency of 6 Hz. The mass of the block is, approximately,

A particle executing linear SHM has velocities $v_{1}$ and $v_{2}$ at distances $x_{1}$ and $x_{2}$ , respectively, from the mean position. The angular velocity of the particle is

A particle executes linear SHM with period 12 s. To traverse a distance equal to half its amplitude from the equilibrium position, it takes

The minimum time taken by a particle in SHM with period T to go from an extreme position to a point half way to the equilibrium position is

The acceleration of a particle performing SHM is 3 $m / s^{2}$ at a distance of 3 cm from the mean position. The periodic time of the motion is

A particle performing linear SHM with a frequency n is confined within limits x = $\pm A$ . Midway between an extremity and the equilibrium position, its speed is

Two spring-and-block oscillators oscillate harmonically with the same amplitude and a constant phase difference of 90°. Their maximum velocities are v and v + x. The value of x is

If the length of a simple pendulum is increased to 4 times its initial length, its frequency of oscillation will