$A$ particle in $S.H.M.$ is described by the displacement function $x(t) = a\cos (\omega t + \theta )$. If the initial $(t = 0)$ position of the particle is $1 \, cm$ and its initial velocity is $\pi \, cm/s$. The angular frequency of the particle is $\pi \, rad/s$,then its amplitude is

Two particles are oscillating along two close parallel straight lines side by side,with the same frequency and amplitudes. They pass each other,moving in opposite directions when their displacement is half of the amplitude. The mean positions of the two particles lie on a straight line perpendicular to the paths of the two particles. The phase difference is

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 performs simple harmonic motion with a period of $2 \ s$. The time taken by the particle to cover a displacement equal to half of its amplitude from the mean position is $\frac{1}{a} \ s$. The value of $a$ to the nearest integer is:

$A$ particle initially at the mean position is executing simple harmonic motion with an angular frequency $\omega = \frac{\pi}{4} \text{ rad s}^{-1}$. The ratio of the distances travelled by the particle in the first second and the second second is:

What is the phase difference between the simple harmonic motions $x = a \sin(\omega t - \alpha)$ and $y = b \cos(\omega t - \alpha)$?