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JEE Main 202125 FebruaryEvening Shift
Exam
JEE Main
Year
2021
Date
25 February
Shift
Evening Shift
Subject
Physics
Chapter
Oscillations

JEE Main 2021 · 25 February · Evening Shift · Physics · Oscillations

The point A moves with a uniform speed along the circumference of a...

The point AA moves with a uniform speed along the circumference of a circle of radius 0.36m0 . 36m and covers 30°30 ° in 0.1s0 . 1s. The perpendicular projection PP from AA on the diameter MNMN represents the simple harmonic motion of PP. The restoration force per unit mass when PP touches MM will be :

The point A moves with a uniform speed along the circumference of a circle of radius 0 . 36m and covers 30 ° in 0 . 1s

Options

  1. A0.49N0 . 49N
  2. B9.87N9 . 87N
  3. C50N50N
  4. D100N100N

Correct answer

B. 9.87N9 . 87N

Step-by-step solution

The point A moves with a uniform speed along the circumference of a circle of radius 0 . 36m and covers 30 ° in 0 . 1s

The point AA covers 30°in0.1s30 ° in 0 . 1s.
Then, we have π6rad⟶0.1s\dfrac{\pi}{6}rad ⟶ 0 . 1s.
Or 1rad⟶0.1π6s1rad ⟶ \dfrac{0 . 1}{\dfrac{\pi}{6}}s
Or 2πin→0.1×6π×2π2\pi in \to \dfrac{0 . 1 \times 6}{\pi} \times 2\pi
Thus, time taken is T=1.2sec.</p><p>WeknowthatT = 1 . 2sec .</p><p>We know that\omega = 2πT<br/>Then,\dfrac{2\pi}{T}<br />Then,\omega = 2π1.2<br/>Restorationforce\dfrac{2\pi}{1 . 2}<br />Restoration force( F ) = {m \omega}^{2}A .

Then Restoration force per unit mass \left(\dfrac{F}{m}\right) = {\omega}^{2}A<br/>(Fm)=(2π1.2)2×0.36<br />\left(\dfrac{F}{m}\right) = {\left(\dfrac{2\pi}{1 . 2}\right)}^{2} \times 0 . 36

≅9.87N≅ 9 . 87N

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