MCQ
The displacement of a particle in $\text{SHM}$ varies according to the relation $\text{x}=4(\cos\pi\text{t}+\sin\pi\text{t})$ The amplitude of the particle is:
  • A
    $-4$
  • B
    $4$
  • $4\sqrt{2}$
  • D
    $8$

Answer

Correct option: C.
$4\sqrt{2}$
Given equation $\text{x(t)}=4(\cos\pi\text{t}+\sin\pi\text{t})$
Now comparing cbove equation with general form $\text{x(t)}=\text{A}\cos\omega\text{t}+\text{B}\sin\omega\text{t}$
We get $A = 4$ and $B = 4$
As, the resultant amplitude for such a equation is
$=\sqrt{\text{A}^2+\text{B}^2}$
$\therefore$ Amplitude $=\sqrt{4^2+4^2}=4\sqrt{2}$

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