Question

An SHM is represented by
$x=5\sqrt{2}(sin2\pi t+cos2\pi t)$
The amplitude of the SHM is

• A. 10 cm
• B. 20 cm
• C. $5\sqrt{2}cm$
• D. 50 cm

Answer 1

Answer: (A)

Here, $x=5\sqrt{2}(sin2\pi t+cos2\pi t)$
$\Rightarrow x=5\sqrt{2}sin2\pi t+5\sqrt{2}cos2\pi t$ ... (i)
The standard equation of simple harmonic motion is given by
$x=A_{1}sin\omega t+A_{2}cos\omega t$... (ii)
Now, comparing Eqs. (i) and (ii), we obtain
$A_1=5\sqrt{2}and{A}_2=5\sqrt{2}$
So, the resuitant amplitude of the motion
$A=\sqrt{A_1^2+A_2^2}=\sqrt{(2\sqrt{2}{)}^2+(5\sqrt{2}{)}^2}$
$=\sqrt{50+50}$
$=10cm$