Frequency of a particle performing $S.H.M.$ is $10 \ Hz$. The particle is suspended from a vertical spring. At the highest point of its oscillation,the spring is unstretched. The maximum speed of the particle is $(g = 10 \ m/s^2)$.

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:

The frequency of oscillation of the spring-mass system shown in the figure is:

Two particles $A$ and $B$ of equal masses are suspended from two massless springs of spring constants $K_A$ and $K_B$ respectively. The maximum velocities of the particles during oscillations are equal. The ratio of the amplitude of '$B$' to that of '$A$' is

$A$ coin is placed on a horizontal platform. The platform performs vertical simple harmonic motion with an angular frequency $\omega$. The amplitude of oscillation is gradually increased. At what condition will the coin first lose contact with the platform?

$A$ particle of mass $0.50 \ kg$ executes simple harmonic motion under force $F = -50 \ (N/m) x$. The time period of oscillation is $\frac{x}{35} \ s$. The value of $x$ is . . . . . (Given $\pi = \frac{22}{7}$)