Q. The amplitude of SHM $y$ $= 2(sin5\pi t + \sqrt{2}cos\pi t)$ is
A) 2
B) $2\sqrt{2}$
C) $4$
D) $2\sqrt{3}$
Explanation
(d) The equation of SHM is
$y = A\sin{(\omega t + \varnothing)}$
or $y = A\ (\sin{\omega t\cos{\varnothing + \cos{\omega t\sin\varnothing}}})$ …(i)
The given expression is
$y = 2(sin5\ \pi t + \sqrt{2}{cos\ }{\pi t})$ …(ii)
$A{cos\ }\varnothing = 2\ \ \ and\ \ \ \ \ \ A{sin\ \ }\varnothing = 2\sqrt{2}$
Squaring and adding, we get
$A = 2\sqrt{3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.