1. The motion of a particle executing simple harmonic motion is described by the displacement function, $x (t) = A cos (ω t + ϕ)$ . If the initial $(t = 0)$ position of the particle is $1 c m$ and its initial velocity is $ω c m / s$ , what are its amplitude and initial phase angle ? The angular frequency of the particle is $π s^{- 1}$ .

2. If instead of the cosine function, we choose the sine function to describe the SHM : $x = B sin (ω t + α)$ , what are the amplitude and initial phase of the particle with the above initial conditions.

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Solution

1. Given, initially $(t = 0)$ the position of the particle (Displacement), $x = 1 c m$
Initial velocity, $v = ω c m / s$
It is given that
$x (t) = A cos (ω t + ϕ)$
$\Rightarrow 1 = A cos (0 + ϕ)$
$A cos ϕ = 1$ .....… $(i)$

$v (t) = \frac{d (x)}{d t} = - A ω sin (ω t + ϕ)$
At $t = 0, v = ω,$ therefore,
$ω = - A ω sin (0 + ϕ)$
$A sin ϕ = - 1$ ......… $(i i)$

From equation $(i)$ and $(i i)$ ,
$A^{2} {sin}^{2} ϕ + A^{2} {cos}^{2} ϕ = 2$
$A^{2} ({sin}^{2} ϕ + {cos}^{2} ϕ) = 2$
$A^{2} = 2$
$A = \sqrt{2} c m$

From equation $(i)$ and $(i i)$
$tan ϕ = - 1$
$∴ ϕ = 3 π / 4, 7 π / 4,$ ...

2. If $S H M$ is,
$x = B sin (ω t + α)$
$1 = B sin (0 + α)$

$\Rightarrow B sin α = 1$ ......... $(i)$

$v (t) = \frac{d (x)}{d t} = B ω cos (ω t + α)$

At $t = 0$ , $v = ω,$ therefore,
$ω = B ω cos (0 + α)$

$B cos α = 1$ .......... $(i i)$

From equation $(i)$ and $(i i)$ ,

$B^{2} {sin}^{2} α + B^{2} {cos}^{2} α = 2$
$B^{2} ({sin}^{2} α + {cos}^{2} α) = 2$
$B^{2} = 2$
$B = \sqrt{2} c m$

From equation $(i)$ and $(i i)$ ,
$tan α = 1$
$α = {tan}^{- 1} (1)$
$∴ α = π / 4, 5 π / 4$ ,...

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