Question

The motion of a particle executing simple harmonic motion is described by the displacement function $x\left(t\right)=A\cos (wt+\varphi )$. If the initial ( t= 0) position of the particle is 1 cm and its initial velocity is w 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 $x=B\sin (wt+\alpha )$ , what are the amplitude and initial phase of particle with the above initial conditions?

Answer 1

intiatlly at $t=0$,

displacement $x=1cm$

initial velocity, $v=\omega cm/sec$.

angular frequency, $\omega =\pi rad/s^{-1}$

its is given that,

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

$1=Acos(wt\times 0=\varphi )=Acos\varphi$

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

now velocity, $v=\frac{dx}{dt}$

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

$Asin\varphi =-1$ . . . . . . . . .(2)

squaring and adding equations (1)and (2) we get,

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

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

$\therefore A=\sqrt{2}cm$

dividing equation (2) and (1) we get

$tan\varphi =-1$

$\therefore \varphi \frac{3\pi }{4},\frac{7\pi }{4},......$

SHM is given as :

$x=Bsin(wt+\alpha )$

putting the given values in this equation, we get:

$1=Bsin[\omega \times 0+\alpha ]=1+1$

$Bsin\alpha =1$ . . . . . . . . . .(3)

velocity, $v=\omega Bcos(wt+\alpha )$

substituting the given values, we get :

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

$Bsin\alpha =1$ . . . . . . . . . . .(4)

squaring and adding equation $3$ and $4$ we get

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

$B=\sqrt{2}cm$

Dividing equation (3) by (4) we get,

$B\frac{sin\alpha }{Bcos\alpha }=\frac{1}{1}$

$tan\alpha =1=tan\frac{\pi }{4}$

$\alpha =\frac{\pi }{4},\frac{5\pi }{4}$