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JEE Main 202528 JanuaryEvening Shift

Exam: JEE Main
Year: 2025
Date: 28 January
Shift: Evening Shift
Subject: Physics
Chapter: Oscillations

JEE Main 2025 · 28 January · Evening Shift · Physics · Oscillations

Given below are two statements. One is labelled as Assertion (A)...

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

x_0

and initial momentum

p_0

is enough to determine the position and momentum at any time

t

for a simple harmonic motion with a given angular frequency

\omega

. Reason (R): The amplitude and phase can be expressed in terms of

x_0

and

\mathrm{p}_0

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

Options

A(A) is false but (R) is true
B(A) is true but (R) is false
CBoth (A) and (R) are true but (R) is NOT the correct explanation of (A)
DBoth $(\mathbf{A})$ and $(\mathbf{R})$ are true and $(\mathbf{R})$ is the correct explanation of $(\mathbf{A})$

Correct answer

D. Both $(\mathbf{A})$ and $(\mathbf{R})$ are true and $(\mathbf{R})$ is the correct explanation of $(\mathbf{A})$

Step-by-step solution

\begin{aligned} & \text{ If we express position } x(t)=A \sin (\omega t+\phi) \\ & \text{ then } x_0=A \sin \phi \\ & v_0=A \omega \cos \phi \\ & \Rightarrow \tan \phi=\frac{\omega x_0}{v_0} \\ & A=\sqrt{x_0^2+\frac{v_0^2}{\omega^2}}\end{aligned}

Hence both position and linear momentum of a particle can be expressed as a function of time if we know initial momentum and position

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