The motion of a particle executing simple harmonic motion is desc… - Vidyadip

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.

✓

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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