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Simple Harmonic Motion question

2021 · 25 Feb · Shift 1 · Q57
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  5. /2021 · 25 Feb · Shift 1 · Q57

Simple Harmonic Motion question

2021 · 25 Feb · Shift 1 · Q57

JEE MainPhysicsSimple Harmonic MotionMCQ+4 / −1
If the time period of a two meter long simple pendulum is 2s, the acceleration due to gravity at the place where pendulum is executing S.H.M. is :
  1. A
    16 m/s2
  2. B
    2 π\piπ 2 ms −-− 2
  3. C
    π\piπ 2 ms −-− 2
  4. D
    9.8 ms −-− 2
View written solutionFree

Correct answer: B

  1. For a simple pendulum, the time period is

T=2πlgT = 2\pi \sqrt{\frac{l}{g}}T=2πgl​​

  1. Given:

l=2 m,T=2 sl = 2\,\text{m}, \qquad T = 2\,\text{s}l=2m,T=2s

Substitute into the formula:

2=2π2g2 = 2\pi \sqrt{\frac{2}{g}}2=2πg2​​

  1. Divide both sides by 2π2\pi2π:

1π=2g\frac{1}{\pi} = \sqrt{\frac{2}{g}}π1​=g2​​

  1. Squaring both sides:

1π2=2g\frac{1}{\pi^2} = \frac{2}{g}π21​=g2​

  1. Rearranging for ggg:

g=2π2 m/s2g = 2\pi^2\,\text{m/s}^2g=2π2m/s2

  1. Compare with options:
  • A: 16 m/s216\,\text{m/s}^216m/s2
  • B: 2π2 m/s22\pi^2\,\text{m/s}^22π2m/s2
  • C: π2 m/s2\pi^2\,\text{m/s}^2π2m/s2
  • D: 9.8 m/s29.8\,\text{m/s}^29.8m/s2

Hence, the correct option is B.

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