Question

The displacement of a particle executing simple harmonic motion is given by
$x=3sin[2\pi t+\frac{\pi }{4}]$ where x is in metres and t is in seconds. The amplitude and maximum speed of the particle is:

• A. $3m,2\pi m{s}^{-1}$
• B. $3m,4\pi m{s}^{-1}$
• C. $3m,6\pi m{s}^{-1}$
• D. $3m,8\pi m{s}^{-1}$

Answer 1

The correct option is C $3m,6\pi m{s}^{-1}$

The given equation of SHM is

$x=3sin[2\pi t+\frac{\pi }{4}]$

Compare the given equation with standard equation of SHM
$x=Asin(\omega t+\varphi )$

we get, $A=3m,\omega =2\pi s^{-1}$

$\therefore Maximumspeed,v_{max}=A\omega =3m\times 2\pi s^{-1}=6\pi m{s}^{-1}$