JEE Main 2022 · 27 Jun · Shift 2 · Physics Q40

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PhysicsSingle correct+4 / -1Question 40 of 90

The equation of a particle executing simple harmonic motion is given by $x=\sin\pi\left(t+\dfrac13\right)$ m. At $t=1$ s, the speed of particle will be

(Given: $\pi=3.14$ ) -

0 cm s

^{-1}

157 cm s

^{-1}

272 cm s

^{-1}

314 cm s

^{-1}

Answer and worked solution

Official answer: B

Velocity. $v=\pi\cos\pi\left(t+\frac13\right)$ ; at $t=1$ , $\cos\frac{4\pi}3=-\frac12$ , so the speed is $\frac\pi2=1.57$ m s $^{-1}=157$ cm s $^{-1}$ .

(B)

Topics:Simple harmonic motion

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JEE Main 2022 · 27 Jun · Shift 2, Physics question 40