Question

Two simple harmonic motions are represented by $y_1=4\sin \left(4\pi t-\frac{\pi }{2}\right)$ and $y_2=3\cos \left(4\pi t\right).$ The resultant amplitude is :

• A. $7$
• B. $1$
• C. $5$
• D. $2+\sqrt{3}$

Answer 1

The correct option is B $1$

Given,

$y_1=4sin(4\pi t-\frac{\pi }{2})$

$y_2=3cos\left(4\pi t\right)=3sin(4\pi t+\frac{\pi }{2})$

$a_1=4m$

$a_2=3m$

${\alpha }_1=-\frac{\pi }{2}$

${\alpha }_2=\frac{\pi }{2}$

${\alpha }_1-{\alpha }_2=-\pi$

The resultant amplitude,

$R=\sqrt{a_1^2+a_2^2+2a_1{a}_{2}cos(-\pi )}$

$R=\sqrt{a_1^2+a_2^2-2a_1{a}_2}$

$R=\sqrt{(a_1-a_2{)}^2}$

$R=(a_1-a_2)$

$R=1m$

The correct option is B.