Question

y(x,t)=0.6/((2x-5t)^2+5) is a wave pulse. find itsmax displacement(amplitude)?

Answer 1

0.12

Answer 2

The given wave pulse is:

$$y(x,t) = \frac{0.6}{(2x-5t)^2 + 5}.$$

The maximum displacement (amplitude) will occur when the denominator is minimum. Since $(2x-5t)^2 \geq 0$ for all $x$ and $t$, the minimum value of the denominator is when:

$$(2x-5t)^2 = 0 \quad \Longrightarrow \quad (2x-5t)^2 + 5 = 5.$$

Thus, the maximum displacement is:

$$y_{\text{max}} = \frac{0.6}{5} = 0.12.$$