A particle initially at the mean position is executing simple harmonic motion with an angular frequency…
A particle initially at the mean position is executing simple harmonic motion with an angular frequency $\frac{\pi}{4} \mathrm{rad} \mathrm{s}^{-1}$. The ratio of the distances travelled by the particle in the first second and second is
$2: 1$
$1: 1$
$(1+\sqrt{3}): 1$
$(1+\sqrt{2}): 1$
Solution
Angular frequency, $\omega=\frac{\pi}{4}$
Displacement of particle executing SHM at $t=1 \mathrm{~s}$
$Y_1=A \sin \frac{\pi}{4}=\frac{A}{\sqrt{2}}$
Displacement of particles executing S H M at $t=2 s$
$Y_2=A \sin \frac{2 \pi}{4}=A$
Distance covered in time $2 \mathrm{~s}, \mathrm{Y}_2-\mathrm{Y}_1=\mathrm{A}-\frac{\mathrm{A}}{\sqrt{2}}$
Ratio $=\frac{Y_1}{Y_2-Y_1}=\frac{\frac{A}{\sqrt{2}}}{A-\frac{A}{\sqrt{2}}}=\frac{1}{\sqrt{2}-1} \times \frac{\sqrt{2}+1}{\sqrt{2}+1}$
Ratio $=(1+\sqrt{2}): 1$