Question

Consider a wave exhibiting simple harmonic oscillation with a period of 6 s. If 4 s after the wave begins it exhibits a displacement of 0.5 m, what is the amplitude of the wave

Answer 1

Answer:

A = 1.0m

Explanation:

In order to calculate the amplitude of the wave you take into account the general formula for a simple harmonic motion, which is given by:

$x=Acos(\omega t)$       (1)

A: amplitude

x: position = 0.5m

t: time = 4s

w: angular frequency of the wave

You calculate the angular frequency of the wave by using the information about the period of the wave:

$\omega=\frac{2\pi}{T}=\frac{2\pi}{6s}=1.047\fra{rad}{s}$

Next, you solve the equation (1) for A, and replace the values of the other parameters:

$A=\frac{x}{cos(\omega t)}=\frac{0.5m}{cos(1.047rad/s(4s))}=-1$

You take the absolute value of the amplitude.

The amplitude of the wave is 1.0 m