Question

In case of super position of waves (at $x=0$),
$y_1=4\sin \left(1026\pi t\right)$ and $y_2=2\sin \left(1014\pi t\right)$

a) the frequency of resulting wave is $510$ Hz
b) the amplitude of resulting wave varies at the frequency of $3$ Hz
c) the frequency of beats is $6$ per second
d) the ratio of maximum to minimum intensity is $9$

The correct statements are

• A. a,d only
• B. b,d only
• C. a, c, d only
• D. a,b,c,d

Answer 1

The correct option is D a,b,c,d

Beat $={\delta }_1-{\delta }_2$$=\frac{{\omega }_1}{2\pi }-\frac{{\omega }_2}{2\pi }$$=\frac{1026\pi }{2\pi }-\frac{1014\pi }{2\pi }$$=6$

$\frac{Imax}{Imin}=\frac{({\Delta }_1+A_2{)}^2}{({\Delta }_1{A}_2{)}^2}=\frac{(4+2{)}^2}{(4-2{)}^2}=\frac{36}{4}=\frac{9}{1}$

$y_1=4sin\left(1026\pi t\right)$
$y_2=2sin\left(1014\pi t\right)$
$y=y_1+y_2$
$=4Sin\left(1026\pi t\right)+2sin\left(1014\pi t\right)$
$=\left(4\sqrt{(\frac{3}{1}{)}^2+cos(12\pi t)}\right)sin\left(1020\pi t\right)$

So, clearly frequency $=\frac{\omega }{2\pi }=\frac{1020\pi }{2\pi }=510Hz$
and amplitude of resulting wave varies at frequency
$\delta =\frac{\omega }{2\pi }=\frac{6\pi }{2\pi }=3Hz$