$A$ simple harmonic oscillator of frequency $1 \ Hz$ has a phase of $1$ radian. By how much should the origin be shifted in time so as to make the phase of the oscillator vanish? ($t$ in seconds).

Two bodies performing $SHM$ have the same amplitude and frequency. Their positions and directions of motion at a certain instant are as shown in the figure. The phase difference between them is

The phase difference between two $SHM$ equations $y_1 = 10 \sin(10\pi t + \frac{\pi}{3})$ and $y_2 = 12 \sin(8\pi t + \frac{\pi}{4})$ at $t = 0.5 \ s$ is:

The periodic time of a particle doing simple harmonic motion is $4 \,s$. The time taken by it to go from its mean position to half the maximum displacement (amplitude) is

$A$ particle executes $SHM$ with a period of $1.2 \, s$ and an amplitude of $8 \, cm$. Find the time it takes to travel $3 \, cm$ from the positive extremity of its oscillation.

$A$ 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