Question

Standing waves are produced by the superposition of two waves, $y_1=0.05\sin (3\pi t-2x)$ and $y_2=0.05\sin (3\pi t+2x)$, where $x$ and $y$ are expressed in metres and $t$ is in seconds. What is the amplitude of a particle at $x=0.5\text{m}$? Given $\cos {57.3}^{\circ }=0.54$.

• A. $2.7\text{cm}$
• B. $5.4\text{cm}$
• C. $8.1\text{cm}$
• D. $10.8\text{cm}$

Answer 1

The correct option is B $5.4\text{cm}$

The resultant displacement is given by
$y=y_1+y_2=0.05\left\{\sin \right(3\pi t-2x)+\sin (3\pi t+2x\left)\right\}$
Using the trigonometric relation
$\sin (\alpha +\beta )+\sin (\alpha -\beta )=2\sin \alpha \cos \beta$,
we have $y=0.1\cos 2x.\sin 3\pi t$

i.e $y=R\sin 3\pi t$
where $R$, the amplitude of standing wave is given by $R=0.1\cos 2x$
When $x=0.5\text{m}$,
$\cos 2x=\cos (2\times 0.5)=\cos \left(1\text{rad}\right)$
$=\cos {57.3}^{\circ }=0.54$

$\therefore$ Amplitude at $x=0.5$ is $R=0.1\times 0.54=0.054\text{m}=5.4\text{cm}$