Question

how to find the amplitude of y = 0.8/[(4x+5t)^2+5]

Answer 1

As per the given question,

$y =\dfrac{ 0.8}{[(4x+5t)^2+5]}$ here y is the displacement about the equilibrium,

So, displacement will be maximum, if denominator will be minimum,

$\Rightarrow [(4x+5t)^2]=0$

$\Rightarrow 4x+5t=0$

$x=-5t/4$

$y=\dfrac{0.8}{5}=0.16$

at t=0,