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Topic Question Set

Q 11 :    

A particle osacillates along the x-axis according to the law, x(t)=x0sin2(t2) where x0=1 m. The kinetic energy (K) of the particle as a function of x is correctly represented by the graph.          [2025]

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(1)

x(t)=x0sin2(t2)=x02(1–cos t)

Clearly x02 is mean position, Particle is oscilating between

Q 12 :    

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 and initial momentum p0 is enough to determine the position and momentum at any ime t for a simple harmonic motion with a given angular frequency ω.

Reason (R) : The ampliitude and phase can be expressed in terms of x0 and p0.

In the light of the above statements, choose the correct answer from the options given below:          [2025]

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

     

  • (A) is false but (R) is true.

     

  • (A) is true but (R) is false.

     

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

     

(4)

x=A sin (ωt+ϕ)

x0=A sin ϕ          ... (i)

p=mAω cos (ωt+ϕ)

p0=mAω cos ϕ          ... (ii)

(ii)/(i)

⇒ tan ϕ=(x0p0)mω

      sin ϕ=x0mω(mωx0)2+p02

From (i),

      A=x0sin ϕ=sin ϕ=(mωx0)2+p02mω

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

Q 13 :    

Which of the following curves possibly represent one-dimensional motion of a particle?         [2025]

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(1)

A. Phase increase with time in SHM, ϕ = kt + C

For example, in SHM, x = A sin ϕ

⇒ Correct

B. In SHM Velocity and displacement are related in elliptical/circular relation

i.e., v2+x2 = constant, it can be 1 D motion

⇒ Correct

C. At same time particle can't have two velocity ⇒ Incorrect.

D. Distance always increases ⇒ Correct

Hencem A, B and D are correct.