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Physics · Simple Harmonic Motion

JEE Main 2025 — 28 January, Evening Shift — Question 56

Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R).

Assertion (A) : Knowing initial position x0\mathrm{x}_{0} and initial momentum p0p_{0} is enough to determine the position and momentum at any time t for a simple harmonic motion with a given angular frequency ω\omega.

Reason (R) : The amplitude and phase can be expressed in terms of x0\mathrm{x}_{0} an p0\mathrm{p}_{0}.

In the light of the above statements, choose the correct answer from the options given below :

  1. Option A:

    Both (A) and (R) are true but (R) is NOT the correct explanation of (A).

  2. Option B:

    (A) is false but (R) is true.

  3. Option C:

    (A) is true but ( R ) is false.

  4. Option D:

    Both (A) and (R) are true and (R) is the correct explanation of (A).

    Correct

Answer: D

Step-by-step solution

x=Asin⁡(ωt+ϕ)\mathrm{x}=\mathrm{A} \sin (\omega \mathrm{t}+\phi)

x0=Asin⁡ϕ\mathrm{x}_{0}=\mathrm{A} \sin \phi

p=mAωcos⁡(ωt+ϕ)\mathrm{p}=\mathrm{mA} \omega \cos (\omega \mathrm{t}+\phi)

p0=mAωcos⁡ϕ\mathrm{p}_{0}=\mathrm{mA} \omega \cos \phi

(2)/(1)⇒tan⁡ϕ=(x0p0)mω(2) /(1) \Rightarrow \tan \phi=\left(\frac{\mathrm{x}_{0}}{\mathrm{p}_{0}}\right) \mathrm{m} \omega

sin⁡ϕ=x0 mω( mxx0)2+p02\sin \phi=\frac{\mathrm{x}_{0} \mathrm{~m} \omega}{\sqrt{\left(\mathrm{~m}_{\mathrm{x}} \mathrm{x}_{0}\right)^{2}+\mathrm{p}_{0}^{2}}}

From (1), A=x0sin⁡ϕ=(mωx0)2+p02mωA=\frac{x_{0}}{\sin \phi}=\frac{\sqrt{\left(m \omega x_{0}\right)^{2}+p_{0}^{2}}}{m \omega}

This means we can explain assertion with the given reason.

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
28 January, Evening Shift
Subject
Physics
Chapter
Simple Harmonic Motion
Topic
Simple Pendulum and Angular SHM