Question

A particle in S.H.M is described by the displacement function $x\left(t\right)=a\cos (\omega t+\theta )$. If the initial (t = 0) position of the particle is 1 cm and its initial velocity is $\pi$ cm/s. The angular frequency of the particle is $\pi rad/s$, then its amplitude

• A. $1cm$
• B. $\sqrt{2}$ cm
• C. $2cm$
• D. $2.5cm$

Answer 1

Answer: (B)

$\sqrt{2}$ cm

$x=a\cos (\omega t+\theta )....\left(i\right)$

and $v=\frac{dx}{dt}=-\omega \sin (\omega t+\theta ).....\left(ii\right)$

Given at $t=0,x=1cm$ and $v=\pi$ and $\omega =\pi$

Putting these values in equation (i) and (ii) we will get $\sin \theta =\frac{-1}{a}$ and $\cos \theta =\frac{1}{a}$

$\Rightarrow {\sin }^2\theta +{\cos }^2\theta ={(-\frac{1}{a})}^2+{\left(\frac{1}{a}\right)}^2\Rightarrow a=\sqrt{2}cm$