The Displacement - time wquation of a particale executing SHM is- x-Asin - t - - At time t-0- position of the particle is x-A2 and it is moving along nagative x - direction- Then the angle - can be-

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Instantaneous Velocity

The Displacem...

Question

The Displacement - time wquation of a particale executing $S H M i s : x = A sin (ω t + ϕ)$ . At time $t = 0$ , position of the particle is $x = \frac{A}{2}$ and it is moving along nagative $x$ - direction. Then the angle $ϕ$ can be:

π

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\frac{π}{6}

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\frac{3 π}{6}

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\frac{5 π}{6}

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Solution

The correct option is D $\frac{5 π}{6}$
Given, $x = A sin (ω t + ϕ)$
at $t = 0, x = \frac{A}{2}$
Therefore,
$\frac{A}{2} = A sin ϕ \Rightarrow ϕ = \frac{π}{6}, \frac{5 π}{6}$
As particle is moving along $- x$ direction, means velocity is negative
Therefore,
$v = \frac{d x}{d t} v = A ω cos (ω t + ϕ)$
At $t = 0$
$v = A w cos ϕ$
Hence, for negative direction of velocity at $t = 0$
$ϕ = \frac{5 π}{6}$

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Instantaneous Velocity

Standard XII Physics

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The Displacement - time wquation of a particale executing SHM is:x=Asin(ωt+ϕ). At time t=0, position of the particle is x=A2 and it is moving along nagative x - direction. Then the angle ϕ can be:

The correct option is D 5π6Given, x=Asin(ωt+ϕ) at t=0,x=A2 Therefore, A2=Asinϕ⇒ϕ=π6,5π6 As particle is moving along −x direction, means velocity is negative Therefore, v=dxdtv=Aωcos(ωt+ϕ) At t=0 v=Awcosϕ Hence, for negative direction of velocity at t=0 ϕ=5π6

The Displacement - time wquation of a particale executing $S H M i s : x = A sin (ω t + ϕ)$ . At time $t = 0$ , position of the particle is $x = \frac{A}{2}$ and it is moving along nagative $x$ - direction. Then the angle $ϕ$ can be: