Question

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

Answer 1

Intially, at t = 0;
Displacement, $x=1cm$
Intial velocity, $v=\omega cm/sec.$
Angular frequency, $\omega =\pi rad/s^{-1}$
It is given that,

$x\left(t\right)=Acos(\omega t+\varphi )$

$1=Acos(\omega \times 0+\varphi )=Acos\varphi$

$Acos\varphi =1$ ...(i)

Velocity, $v=dx/dt$

$\omega =-A\omega sin(\omega t+\varphi )$

$1=-Asin(\omega \times 0+\varphi )=-Asin\varphi$

$Asin\varphi =-1$ ...(ii)

Squaring and adding equations (i) and (ii), we get:

$A^2(si{n}^2\varphi +co{s}^2\varphi )=1+1$
$A^2=2$

$\therefore A=\sqrt{2}cm$
Dividing equation (ii) by equation (i), we get:

$tan\varphi =-1$
$\therefore \varphi =3\pi /4,7\pi /4$
SHM is given as:

$x=Bsin(\omega t+\alpha )$
Putting the given values in this equation, we get:

$1=Bsin(\omega \times 0+\alpha ]=1+1$
$Bsin\alpha =1$ ...(iii)
Velocity, $v=\omega cos(\omega t+\alpha )$
Substituting the given values, we get:

$\pi =\pi Bcos\alpha$

$Bcos\alpha =1$ ...(iv)
Squaring and adding equations(iii) and (iv), we get:

$B^2[si{n}^2\alpha +co{s}^2\alpha ]=1+1$
$B^2=2$

$\therefore B=\sqrt{2}$
Dividing equation (iii) by equation (iv), we get:

$Bsin\alpha /Bcos\alpha =1/1$
$tan\alpha =1=tan\pi /4$
$\therefore \alpha =\pi /4,5\pi /4,$......