MCQEasyJEE 2025Simple Harmonic Motion (SHM)

JEE Physics 2025 Question with Solution

Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): Knowing initial position x0x_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 x0x_0 and p0p_0. In the light of the above statements, choose the correct answer from the options given below:

  • A

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

  • B

    (A) is false but (R) is true

  • C

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

  • D

    (A) is true but (R) is false

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: Initial position x0x_0 and initial momentum p0p_0 for a simple harmonic motion with given angular frequency ω\omega.

Find: Whether Assertion (A) and Reason (R) are true, and whether (R) correctly explains (A).

For SHM, the displacement and momentum may be written as

x(t)=Acos(ωt+ϕ)x(t) = A \cos(\omega t + \phi)

and

p(t)=mdxdt=mωAsin(ωt+ϕ)p(t) = m\frac{dx}{dt} = -m\omega A \sin(\omega t + \phi)

At t=0t = 0,

x0=Acosϕ,p0=mωAsinϕx_0 = A \cos \phi, \quad p_0 = -m\omega A \sin \phi

Squaring and adding,

x02+(p0mω)2=A2x_0^2 + \left(\frac{p_0}{m\omega}\right)^2 = A^2

So,

A=x02+(p0mω)2A = \sqrt{x_0^2 + \left(\frac{p_0}{m\omega}\right)^2}

Also,

tanϕ=p0mωx0\tan \phi = -\frac{p_0}{m\omega x_0}

Therefore, both the amplitude AA and phase ϕ\phi can be determined from x0x_0 and p0p_0. Once AA and ϕ\phi are known, x(t)x(t) and p(t)p(t) are completely determined for any time tt.

Hence, Assertion (A) is true, Reason (R) is true, and Reason (R) correctly explains Assertion (A).

Therefore, the correct option is A.

Step-by-step derivation

Given: Assertion (A): Knowing the initial position x0x_0 and initial momentum p0p_0 is enough to determine the position and momentum at any time tt for a simple harmonic motion with given angular frequency ω\omega.

Reason (R): The amplitude and phase can be expressed in terms of x0x_0 and p0p_0.

Find: The correct option.

Step 1: Write the standard SHM form:

x(t)=Acos(ωt+ϕ)x(t) = A \cos(\omega t + \phi) p(t)=mωAsin(ωt+ϕ)p(t) = -m\omega A \sin(\omega t + \phi)

Step 2: Apply the initial condition t=0t = 0:

x0=Acosϕx_0 = A \cos \phi p0=mωAsinϕp_0 = -m\omega A \sin \phi

Step 3: Determine the amplitude by squaring and adding:

x02+(p0mω)2=A2x_0^2 + \left(\frac{p_0}{m\omega}\right)^2 = A^2

Thus,

A=x02+(p0mω)2A = \sqrt{x_0^2 + \left(\frac{p_0}{m\omega}\right)^2}

Step 4: Determine the phase:

tanϕ=p0mωx0\tan \phi = -\frac{p_0}{m\omega x_0}

So ϕ\phi is also determined by x0x_0 and p0p_0.

Step 5: Since the complete SHM is specified by amplitude AA and phase ϕ\phi, knowing x0x_0 and p0p_0 is sufficient to determine both x(t)x(t) and p(t)p(t) at any time tt.

Therefore, both (A) and (R) are true, and (R) is the correct explanation of (A).

The correct option is A.

Common mistakes

  • Assuming that initial position x0x_0 alone is enough to fix the motion. This is wrong because SHM also depends on the initial momentum or velocity. Use both x0x_0 and p0p_0 to determine amplitude and phase.

  • Thinking that amplitude is always equal to the initial displacement. This is incorrect unless the particle starts from an extreme position. Instead, calculate AA using both x0x_0 and p0p_0.

  • Using the SHM form without accounting for phase. This leads to incomplete determination of motion. The correct approach is to find both AA and ϕ\phi from the initial conditions.

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