Verified Solution • 2025–26 Session moderate

QJEEM25Ph40c5: JEE Main 2025 Physics - Class 11 JEEM25Ph40c5 Solution: Given below are two statements. One is labelled as Assertion…

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 tt 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 :

Multiple Choice Options

  • A (A) is true but (R) is false
  • B Both (A) and (R) are true but (R) is NOT the correct explanation of (A)
  • C (A) is false but (R) is true
  • D Both (A) and (R) are true and (R) is the correct explanation of (A)

Step-by-Step Verified Solution

amp;x=Asin⁡(ωt+ϕ)amp;x0=Asin⁡ϕ.... (1)amp;p=mAωcos⁡(ωt+ϕ)amp;p0=mAωcos⁡ϕ.... (2)\begin{aligned}& \mathrm{x}=\mathrm{A} \sin (\omega \mathrm{t}+\phi) \\& \mathrm{x}_0=\mathrm{A} \sin \phi \quad\text{.... (1)}\\& \mathrm{p}=\mathrm{mA} \omega \cos (\omega \mathrm{t}+\phi) \\& \mathrm{p}_0=\mathrm{mA} \omega \cos \phi \quad\text{.... (2)}\end{aligned}

(2)/(1)⇒tan⁡ϕ=(x0p0)mω(2) /(1) \Rightarrow \tan \phi=\left(\frac{x_0}{p_0}\right) m \omega

sin⁡ϕ=x0 mω( mωx0)2+p02\sin \phi=\frac{\mathrm{x}_0 \mathrm{~m} \omega}{\sqrt{\left(\mathrm{~m} \omega \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.

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