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QJEEM20Ph6eb2: JEE Main 2020 Physics - Class 11 JEEM20Ph6eb2 Solution: The displacement time graph of a particle executing S.H.M is…

The displacement time graph of a particle executing S.H.M is given in figure :
(sketch is schematic and not to scale) JEE Main 2020 (Online) 2nd September Evening Slot Physics - Simple Harmonic Motion Question 103 English
Which of the following statements is/are true for this motion?
(A) The force is zero at t = 3T4{{3T} \over 4}
(B) The acceleration is maximum at t = T
(C) The speed is maximum at t = T4{{T} \over 4}
(D) The P.E. is equal to K.E. of the oscillation at t = T2{{T} \over 2}

Multiple Choice Options

  • A (B), (C) and (D)
  • B (A), (B) and (C)
  • C (A) and (D)
  • D (A), (B) and (D)

Step-by-Step Verified Solution

(A) F = ma and a = −ω2x-\omega^{2} x

At 3T4\frac{3T}{4} displacement zero (x = 0), so a = 0

∴\therefore F = 0

(B) at t = T, Particle is at extreme.

displacement (x) = A ⇒\Rightarrow x maximum,

So acceleration is maximum.

(C) V = ωA2−x2\omega \sqrt{A^{2}-x^{2}}

at t = T4\frac{T}{4} , x = 0

∴\therefore V = A ω\omega = Vmax

(D) KE = PE

12k(A2−x2)=12kx2\frac{1}{2} k\left( A^{2}-x^{2}\right) =\frac{1}{2} kx^{2}

⇒\Rightarrow A2 – x2 = x2

⇒\Rightarrow A2 = 2x2

A = 2\sqrt{2} x

⇒\Rightarrow x = A2\frac{A}{\sqrt{2} } = A cos ω\omega t

⇒\Rightarrow cos ω\omega t = 12\frac{1}{\sqrt{2} }

⇒\Rightarrow ω\omega t = π4\frac{\pi }{4}

⇒\Rightarrow 2πT⋅t=π4\frac{2\pi }{T} \cdot t=\frac{\pi }{4}

⇒\Rightarrow t = T8\frac{T}{8}

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