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Q.
The displacement (x) of a particle as a function of time (t) is given by; $x=a\, \sin (bt+c)$ where a, b and c are constants of motion. Choose the correct statements from the following.
The motion of the particle is simple harmonic.
The displacement at time $t$ is $x=a \sin (b t+c)$
Therefore,displacement at time $(t+(2 \pi / b))$ is
$x$ at $\left(t+\frac{2 \pi}{b}\right)=a \sin \left[b\left(t+\frac{2 \pi}{b}\right)+c\right]$
$=a \sin [b t+c+2 \pi]=a \sin (b t+c)=x($ at time $t)$
Hence, statement (a) is correct.
Statement (b) is also correct since the same displacement is recovered after a time interval of $(2 \pi / b )$.
Statement (c) is correct because the velocity is zero when the displacement $=\pm$ the amplitude, i.e at the extreme ends of the motion.
Statement (d) is incorrect, the acceleration is maximum (in magnitude ) at $x =\pm A .$