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Saturday, 31 October

JEE Main 2022 · PhysicsMultiple choiceSingle correctEasyCalculation

JEE Main 27 June 2022, Shift 2, Physics Q40

Question 40 of 90 in this shift, Physics question 10 of 30, Section A.

The equation of a particle executing simple harmonic motion is given by x=sin⁡π(t+13)x = \sin \pi\left(t + \frac{1}{3}\right) m. At tt = 1s, the speed of particle will be (Given: π\pi = 3.14) -
  1. (1)0 cm s−1^{-1}
  2. (2)157 cm s−1^{-1}Official answer
  3. (3)272 cm s−1^{-1}
  4. (4)314 cm s−1^{-1}

Official answer

Option 2

NTA final key.