$A$ particle executes $SHM$ with amplitude $0.2 \,m$ and time period $24 \,s$. The time required for it to move from the mean position to a point $0.1 \,m$ from the mean position is (in $\,s$)

$A$ particle executes simple harmonic oscillation with an amplitude $a$. The period of oscillation is $T$. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is:

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 is performing simple harmonic motion along the $x$-axis with an amplitude of $4 \, cm$ and a time period of $1.2 \, s$. The minimum time taken by the particle to move from $x = 2 \, cm$ to $x = +4 \, cm$ and back again is given by .... $s$.

The phase of a particle executing simple harmonic motion is $\frac{\pi}{2}$ when it has:

The minimum phase difference between two simple harmonic motions $x_1 = \frac{1}{\sqrt{2}} \sin \omega t + \frac{1}{\sqrt{2}} \cos \omega t$ and $x_2 = \sin \omega t + \cos \omega t$ is $[\sin \frac{\pi}{4} = \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}]$