Question

for the simple harmonic motion d = 9cos (pi//2t), what is the maximum displacement from the equilibrium position?

Answer 1

Y = A cos (B(x - C) + D

A gives the amplitude, which is the maximum displacement, so it's 9.

Answer 2

Answer:

The maximum displacement from the equilibrium position is 9.

Step-by-step explanation:

The given simple harmonic motion is:

$d=9cos(\frac{\pi}{2}t)$

Differentiating above equation with respect to t, we get

$d'=-9sin(\frac{\pi}{2}t)(\frac{\pi}{2})$

⇒$d'=-\frac{9{\pi}}{2}sin(\frac{\pi}{2}t)$

Again differentiating the above equation with respect to t, we have

$d''=-\frac{9({\pi})^{2}}{4}cos(\frac{\pi}{2}t)$

Now, $d'=0$

⇒$-\frac{9{\pi}}{2}sin(\frac{\pi}{2}t)=0$

⇒$sin(\frac{\pi}{2}t)=0$

⇒$\frac{\pi}{2}t=0$

⇒$t=0$

Substituting the value of t=0 in d, we get

⇒$d=9cos(\frac{\pi}{2}(0))$

⇒$d=9cos(0)$

⇒$d=9(1)$

⇒$d=9$

Therefore, the maximum displacement from the equilibrium position is d= 9.