The displacement of a simple harmonic oscillator after $3 \; s$ starting from its mean position is equal to half of its amplitude. The time period of the harmonic motion is $\dots \; s$.

$A$ simple harmonic oscillator has an amplitude $A$ and time period $6 \pi \text{ s}$. Assuming the oscillation starts from its mean position,the time required by it to travel from $x=A$ to $x=\frac{\sqrt{3}}{2} A$ will be $\frac{\pi}{x} \text{ s}$,where $x=$ . . . . . . .

The displacement-time equation of a particle executing $SHM$ is $x = A \sin(\omega t + \phi)$. At time $t = 0$,the position of the particle is $x = A/2$ and it is moving along the negative $x$-direction. Then the phase angle $\phi$ is:

$A$ particle executes simple harmonic motion between $x = -A$ and $x = +A$. It starts from $x = 0$ and moves in the $+x$ direction. The time taken for it to move from $x = 0$ to $x = \frac{A}{2}$ is $T_1$,and the time taken to move from $x = \frac{A}{2}$ to $x = \frac{A}{\sqrt{2}}$ is $T_2$. Then:

Consider a pair of identical pendulums,which oscillate with equal amplitude independently such that when one pendulum is at its extreme position making an angle of $2^o$ to the right with the vertical,the other pendulum makes an angle of $1^o$ to the left of the vertical. What is the phase difference between the pendulums?

$A$ particle performs $SHM$ about $x = 0$ such that at $t = 0$ it is at $x = 0$ and moving towards the positive extreme. The time taken by it to go from $x = 0$ to $x = \frac{A}{2}$ is ..... times the time taken to go from $x = \frac{A}{2}$ to $A$. The most suitable option for the blank space is