Question

The equation of a simple harmonic motion is given by $y=3\sin (50t-x)$, where $x$ and $y$ are in metres and $t$ is in seconds, the maximum particle velocity in $m{s}^{-1}$ is

• A. 3
• B. 50
• C. 150
• D. 25

Answer 1

Answer: (C)

The equation of a simple harmonic motion is given by
$y=3\sin (50t-x)$
By comparing Eq. (i) with general equation of a simple harmonic motion $y=A\sin (\omega t+\varphi )$, we get
Amplitude $A=3m$
angular frequency, $\omega =50Hz$
$\therefore$ Maximum particle velocity, $v_{\max }=A\omega =3\times 50$
$=150m{s}^{-1}$