The equation of a particle executing simple harmonic motion is given by x=sinπt+13m.At t=1 s, the speed of particle will be(Given :π=3.14 )

x=sinπt+13x=sinπt+π3V=dxdt=cosπt+π3π =-π×12=157 cm/s

The equation of a particle executing simple harmonic motion is given by x=sinπt+13m.At t=1 s, the speed of particle will be(Given :π=3.14 )

x=sinπt+13x=sinπt+π3V=dxdt=cosπt+π3π =-π×12=157 cm/s

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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 $: π = 3.14$ )

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$0 c m s^{- 1}$

$157 c m s^{- 1}$

$272 c m s^{- 1}$

$314 c m s^{- 1}$

answer is B.

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Detailed Solution

$\begin{array}{rcl} x & = & \sin π (t + \frac{1}{3}) \\ x & = & \sin (π t + \frac{π}{3}) \\ V & = & \frac{dx}{dt} = \cos (π t + \frac{π}{3}) π \\ = & - π \times \frac{1}{2} = 157 cm / s \end{array}$

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