Question

A particle in SHM is described by the displacement function: $x\left(t\right)=Acos(\omega t+\varphi ).$ If the initial $(t=0)$ position of the particle is $1$ cm and its initial velocity is $\omega$ cm/sec , what is the initial phase angle ?

• A. $\frac{3\pi }{4}$
• B. $\frac{7\pi }{4}$
• C. $\frac{7\pi }{2}$
• D. $\frac{\pi }{4}$

Answer 1

Answer: (A), (B)

The correct options are
A $\frac{3\pi }{4}$
B $\frac{7\pi }{4}$

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

$⟹v\left(t\right)=\frac{dx\left(t\right)}{dt}$

$=-\omega Asin(\omega t+\varphi )$

Initial position=$x(t=0)=Acos\varphi =1cm$

Initial velocity=$v(t=0)=-A\omega sin\varphi =\omega cm/s$

Hence, $\varphi =ta{n}^{-1}(-1)=\frac{7\pi }{4}$