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Oscillations — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsOscillations5 Marks
Question
The motion of a particle executing simple harmonic motion is described by the displacement function,$\text{x(t)}=\text{A}\cos(\omega\text{t}+\phi).$
If the initial (t = 0) position of the particle is 1cm and its initial velocity is $\omega\text{ cm/s,}$ what are its amplitude and initial phase angle? The angular frequency of the particle is $\pi\text{s}^{-1}.$ If instead of the cosine function, we choose the sine function to describe the SHM: $\text{x}=\text{B}\sin(\omega\text{t}+\alpha),$ what are the amplitude and initial phase of the particle with the above initial conditions.
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Answer

Initially, at t = 0: Displacement, x = 1cm Initial velocity, $\text{v}=\omega\text{ cm/sec.}$ Angular frequency, $\omega=\pi\text{ rad/s}^{-1}$ It is given that: $\text{x(t)}=\text{A}\cos(\omega\text{t}+\phi)$ $1=\text{A}\cos(\omega\times0\times\phi)$ $\text{A}\cos\phi=1\ ........(1)$ $\text{Velocity, }\upsilon=\frac{\text{dx}}{\text{dt}}$ $\omega=-\text{A}\omega\sin(\omega\text{t}+\phi)$ $1=-\text{A}\sin(\omega\times0+\phi)=-\text{A}\sin\phi$ $\text{A}\sin\phi=-1\ ....(2)$ Squaring and adding equations (1) and (2), we get: $\text{A}^2(\sin^2\phi+\cos^2\phi)=1+1$ $\text{A}^2=2$ $\therefore\ \text{A}=\sqrt{2}\ \text{cm}$ Dividing equation (2) by equation (1), we get: $\tan\phi=-1$ $\therefore\ \phi=\frac{3\pi}{4},\frac{7\pi}{4},....$ SHM is given as: $\text{x}=\text{B}\sin(\omega\text{t}+\alpha)$Putting the given values in this equation, we get:
$1=\text{B}\sin[\omega\times0+\alpha]$ $\text{B}\sin\alpha=1\ .....(3)$ Velocity, $\upsilon=\omega\text{B}\cos(\omega\text{t}+\alpha)$ Substituting the given values, we get: $\pi=\pi\text{B}\sin\alpha$ $\text{B}\sin\alpha=1\ .....(4)$ Squaring and adding equations (3) and (4), we get:$\text{B}^2[\sin^2\alpha+\cos^2\alpha]=1+1$
$\text{B}^2=2$
$\therefore\ \text{B}=\sqrt{2}\text{ cm}$
Dividing equation (3) by equation (4), we get: $\frac{\text{B}\sin\alpha}{\text{B}\cos\alpha}=\frac{1}{1}$ $\tan\alpha=1=\frac{\tan\alpha}{4}$ $\therefore\ \alpha=\frac{\pi}{4},\frac{5\pi}{4}, .....$

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