Question

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

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

Answer 1

Answer: (C)

The given equation of $SHM$ is
$x = 3 \,sin \left(2\pi t+\frac{\pi}{4}\right)$
Compare the given equation with standard equation of $SHM$
$x = A\, sin \left(\omega t+\phi\right)$
we get, $A = 3\,m, \omega = 2\pi\,s^{-1}$
$\therefore $ Maximum speed, $v_{\text{max}} = A\omega = 3\,m \times 2 \pi \, s^{-1} $
$ = 6 \pi \,m\,s^{-1}$