Question

An object in simple harmonic motion obeys the following position versus time equation: y = (0.50 m) sin (π/2 t). What is the amplitude of vibration and its period of vibration?

Answer 1

Hi there!

The Amplitude is the 'A' value of the following SHM equation:

$x(t) = Asin(\omega t)$

Thus, the amplitude of this object:

A = 0.50 m

The period of vibration is equivalent to:

$T = \frac{2\pi}{\omega}$

In this instance:

$\omega = \frac{\pi}{2}$

Therefore:

$T = \frac{2\pi}{\frac{\pi}{2}} = \frac{4\pi}{\pi} = \boxed{\text{ 4 seconds.}}$