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Home » The motion of a particle executing simple harmonic motion is described by the displacement function, If the initial (t ) position of the particle is and its initial velocity is , what are its amplitude and initial phase angle? The angular frequency of the particle is . If instead of the cosine function, we choose the sine function to describe the SHM: , what are the amplitude and initial phase of the particle with the above initial conditions. Solution:
The motion of a particle executing simple harmonic motion is described by the displacement function, x(t)=A \cos (\omega t+\varphi) If the initial (t =0 ) position of the particle is 1 \mathrm{~cm} and its initial velocity is \omega \mathrm{cm} / \mathrm{s}, what are its amplitude and initial phase angle? The angular frequency of the particle is \pi \mathrm{s}^{-1}. If instead of the cosine function, we choose the sine function to describe the SHM: x=B \sin (w t+a), what are the amplitude and initial phase of the particle with the above initial conditions. Solution:
CBSE | Class 11 | NCERT | Oscillations | Physics
The motion of a particle executing simple harmonic motion is described by the displacement function, x(t)=A \cos (\omega t+\varphi) If the initial (t =0 ) position of the particle is 1 \mathrm{~cm} and its initial velocity is \omega \mathrm{cm} / \mathrm{s}, what are its amplitude and initial phase angle? The angular frequency of the particle is \pi \mathrm{s}^{-1}. If instead of the cosine function, we choose the sine function to describe the SHM: x=B \sin (w t+a), what are the amplitude and initial phase of the particle with the above initial conditions. Solution:

At positlon, t = 0,

The given function is x(t)=A \cos (\omega t+\phi).....(1)

\begin{array}{l} 1=A \cos (\omega \times 0+\phi)=A \cos \phi \\ A \cos \phi=1 \end{array}

Differentiating equation (1) with respect to t, we get,

Velocity, \mathrm{V}=\mathrm{dx} / \mathrm{dt}

\begin{array}{l} v=-\mathrm{A} \omega \sin (\omega t+\phi) \\ \mathrm{t}=0 \text { and } \mathrm{v}=\omega \\ 1=-\mathrm{A} \sin (\omega \times 0+\phi)=-\mathrm{A} \sin \phi \\ \mathrm{A} \sin \phi=-1 \end{array}

On Squaring and adding equations (2) and (4), we get,

\begin{array}{l} A^{2}\left(\sin ^{2} \phi+\cos ^{2} \phi\right)=1+1 \\ A=\sqrt{2} \mathrm{~cm} \end{array}

Dividing equation (4) by equation (2),
(A \sin \phi / A \cos \phi)=-1 / 1
\tan \phi=-1

\begin{array}{l} \Rightarrow \phi=3 \pi / 4,7 \pi / 4 \\ \text { If sine function is used } \\ \mathrm{x}=\mathrm{B} \sin (\omega \mathrm{t}+\mathrm{a}) \\ \text { At } \mathrm{t}=0, \mathrm{x}=1 \text { and } \mathrm{v}=\omega \text { we get } \\ 1=\mathrm{Bsin}(\omega \times 0+\alpha]=1+1 \\ \mathrm{~B} \text { sina }=1-(5) \\ \text { Velocity, } \mathrm{v}=\mathrm{dx} / \mathrm{dt}=\mathrm{B} \omega \cos (\omega t+\alpha) \\ \text { taking } \mathrm{v}=\omega \\ 1=\mathrm{Bcos}(\omega(0)+\mathrm{a})=\mathrm{B} \cos \alpha-(6) \end{array}

Squaring and adding equations(5) and (6), we get:

\begin{array}{l} B^{2}\left[\sin ^{2} \alpha+\cos ^{2} \alpha\right]=1+1 \\ B^{2}=2 \\ B=\sqrt{2} \mathrm{~cm} \end{array}

Dividing equation (5) by equation (6), we get: B \sin \alpha / B \cos \alpha=1 / 1
\tan \alpha=1

As a result, \alpha=\pi / 4,5 \pi / 4, \ldots \ldots

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