What is the ratio between the distance travelled by the oscillator in one time period and the amplitude?

$A$ simple harmonic wave having an amplitude $a$ and time period $T$ is represented by the equation $y = 5 \sin \pi (t + 4) \ m$. Then the value of amplitude $(a)$ in $(m)$ and time period $(T)$ in seconds are:

$SHM$ is associated with what kind of motion?

$A$ particle is executing simple harmonic motion with time period $2 \ s$ and amplitude $1 \ cm$. If $D$ and $d$ are the total distance and displacement covered by the particle in $12.5 \ s$,then $\frac{D}{d}$ is:

The motion of a particle executing simple harmonic motion is described by the displacement function,$x(t) = A \cos (\omega t + \phi)$. If the initial $(t = 0)$ position of the particle is $1 \; cm$ and its initial velocity is $\omega \; cm/s$,what are its amplitude and initial phase angle? The angular frequency of the particle is $\pi \; s^{-1}$. If instead of the cosine function,we choose the sine function to describe the $SHM$: $x = B \sin (\omega t + \alpha)$,what are the amplitude and initial phase of the particle with the above initial conditions?

$A$ particle is executing simple harmonic motion $(SHM)$ with a time period $T = 4 \ s$. At $t = 0$,the particle is at a position where its angular position is $45^\circ$ with the $x$-axis. Find the displacement equation of the particle along the $x$-axis.