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)

3 m, 6$\pi$ m s–1

Answer 2

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

$x = A\sin(\omega t + \varphi)$

We get

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

$\therefore$ Maximum speed ${v - 1}_{\max}$

$= 6\pi ms^{- 1}$