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Physics · Simple Harmonic Motion

JEE Main 2025 — 24 January, Morning Shift — Question 59

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 D/d is :-

  1. Option A:

    15/4

  2. Option B:

    25

    Correct
  3. Option C:

    10

  4. Option D:

    16/5

Answer: B

Step-by-step solution

A=1 cm\mathrm{A}=1 \mathrm{~cm}

n=12.52=6.25\mathrm{n}=\frac{12.5}{2}=6.25 cycles

∴D=4×6+1=25\therefore \mathrm{D}=4 \times 6+1=25

d=1\mathrm{d}=1

Dd=25\frac{\mathrm{D}}{\mathrm{d}}=25

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
24 January, Morning Shift
Subject
Physics
Chapter
Simple Harmonic Motion
Topic
Kinematics of SHM, Phase and Energy in SHM
A particle is executing simple harmonic motion with time period 2 s… | JEE Main 2025 PYQ with Solution · DhiX AI