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

x = s i n π ( t + 1 3 ) v = d x d t = c o s [ π ( t + 1 3 ) ] * π = c o s ( π + π 3 ) * π = − c o s π 3 * ( π ) = − π 2 = − 1 . 5 7 m / s e c Speed = 157 cm/sec

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

(Given : p = 3.14)

Option 1 - 0 cm s-1

Option 2 - 157 cm s-1

Option 3 - 272 cm s-1

Option 4 - 314 cm s-1

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Answered by

Payal Gupta | Contributor-Level 10

6 months ago

Correct Option - 2

Detailed Solution:

$x = s i n π (t + \frac{1}{3})$

$v = \frac{d x}{d t} = c o s [π (t + \frac{1}{3})] * π$

$= c o s (π + \frac{π}{3}) * π$

$= - c o s \frac{π}{3} * (π)$

$= - \frac{π}{2} = - 1.57 m / s e c$

Speed = 157 cm/sec

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