The equation of a particle executing simple harmonic motion is given by x=sinπt+1/3 m. At t=1 s, the speed of particle will be (Given: π=3.14)

Correct answer: (A) 157 cm s^(-1). Given, x=sinπt+(π)/3 Velocity v=dx/dt=πcosπt+(π)/3 Now, at t=1 s, v=πcosπ+(π)/3 v=(-π)/2=(-3.14)/2=-1.57 m s^(-1)⇒v=157 cm s^(-1)

Q70 / 90 · JEE Main 2022 (27 Jun Shift 2) · Physics

The equation of a particle executing simple harmonic motion is given by $x = \sin π (t + \frac{1}{3}) m$ . At $t = 1 s$ , the speed of particle will be (Given: $π = 3.14$ )

Correct answer
A
$157 cm s^{- 1}$
B
$0 cm s^{- 1}$
C
$272 cm s^{- 1}$
D
$314 cm s^{- 1}$

Solution

Given, $x = \sin (π t + \frac{π}{3})$

Velocity $v = \frac{d x}{d t} = π \cos (π t + \frac{π}{3})$

Now, at $t = 1 s$ , $v = π \cos (π + \frac{π}{3})$

$v = \frac{- π}{2} = \frac{- 3.14}{2} = - 1.57 m s^{- 1} \Rightarrow |v| = 157 cm s^{- 1}$

Hence, the correct answer is option: 'A'.

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Solution

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The equation of a particle executing simple harmonic motion is given by $x = \sin π (t + \frac{1}{3}) m$ . At $t = 1 s$ , the speed of particle will be (Given: $π = 3.14$ )