Q. 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 $\frac Dd$ is
A) $\frac{16}5$
B) $10$
C) $\frac{15}4$
D) $25$
Explanation
[Solution by Dr. Abubacker] $12.5\,\text{s}=6.25\,T$. Distance covered in one period is $4A$, so $D=6.25\times4\times1=25$ cm. After $6$ full periods plus a quarter period (starting from the mean position) the displacement is $d=A=1$ cm. $\frac Dd=25$. Answer (4).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.