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

Question

The equation of a particle executing simple harmonic motion is given by x= sin π(t+(1/3)) m. At t =1 s, the speed of particle will be (Given: π=3.14 ) (A) 0 cm s -1 (B) 157 cm s -1 (C) 272 cm s -1 (D) 314 cm s -1

Tardigrade Admin · Accepted Answer

x = sin π( t +(1/3)) x = sin (π t +(π/3)) V =( dx / dt )= cos (π t +(π/3)) π =-π × (1/2)=157 cm / s

Q. 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 : $π = 3.14$ )

$0 c m s^{- 1}$

$157 c m s^{- 1}$

$272 c m s^{- 1}$

$314 c m s^{- 1}$

$x = sin π (t + \frac{1}{3})$
$x = sin (π t + \frac{π}{3})$
$V = \frac{d x}{d t} = cos (π t + \frac{π}{3}) π$
$= - π \times \frac{1}{2} = 157 c m / s$

Q. The equation of a particle executing simple harmonic motion is given by x=sinπ(t+31​)m. At t=1s, the speed of particle will be (Given : π=3.14 )

0cms−1

157cms−1

272cms−1

314cms−1