$A$ particle is performing a linear simple harmonic motion of amplitude $A$. When it is midway between its mean and extreme position,the magnitudes of its velocity and acceleration are equal. What is the periodic time of the motion?

$A$ body executes simple harmonic motion under the action of a force $F_1$ with a time period $T_1 = \frac{4}{5} \ s$. If the force is changed to $F_2$,it executes $SHM$ with a time period $T_2 = \frac{3}{5} \ s$. If both the forces $F_1$ and $F_2$ act simultaneously in the same direction on the body,what will be its new time period in seconds?

$A$ body executes $SHM$ under the influence of one force and has a period $T_1 \, s$. The same body executes $SHM$ with period $T_2 \, s$ when under the influence of another force. When both forces act simultaneously and in the same direction,the time period of the same body is:

$A$ disc of radius $R$ and mass $M$ is pivoted at the rim and is set for small oscillations. If a simple pendulum has to have the same period as that of the disc,the length of the simple pendulum should be

$A$ light hollow cube of side length $10 \ cm$ and mass $10 \ g$ is floating in water. It is pushed down and released to execute simple harmonic oscillations. The time period of oscillations is $y \pi \times 10^{-2} \ s$,where the value of $y$ is (Acceleration due to gravity,$g = 10 \ m/s^2$,density of water $\rho = 10^3 \ kg/m^3$)

$A$ uniform rod of length $L$ and mass $M$ is pivoted at the centre. Its two ends are attached to two springs of equal spring constants $k$. The springs are fixed to rigid supports as shown in the figure,and the rod is free to oscillate in the horizontal plane. The rod is gently pushed through a small angle $\theta$ in one direction and released. The frequency of oscillation is