The amplitude and the periodic time of a $S.H.M.$ are $5 \, cm$ and $6 \, s$ respectively. At a distance of $2.5 \, cm$ away from the mean position,the phase will be

$A$ 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:

$A$ particle performing linear $S.H.M.$ has a period of $8 \ s$. At time $t=0$,it is at the mean position. The ratio of the distances travelled by the particle in the $1^{st}$ and $2^{nd}$ second is $(\cos 45^{\circ} = 1/\sqrt{2})$.

Distance travelled by a particle in $\text{SHM}$ when its phase changes from $\frac{\pi}{6}$ to $\frac{5 \pi}{6}$ is:

The initial phase of a body executing $SHM$ is $\frac{\pi}{4}$. What will be its phase at the end of $10$ oscillations?

If $x = a \sin \left( \omega t + \frac{\pi}{6} \right)$ and $x' = a \cos \omega t$,then what is the phase difference between the two waves?