Question

Two transverse waves passing through a region are represented by $y_1=\left(2\text{cm}\right)\sin \left[\right(2\pi {\text{cm}}^{-1})x-(100\pi {\text{s}}^{-1}\left)t\right]$ and $y_2=\left(3\text{cm}\right)\sin \left[\right(\pi {\text{cm}}^{-1})x-(200\pi {\text{s}}^{-1}\left)t\right]$.
Find the net displacement of the particle at $x=4.5\text{cm}$ and time $t=5\text{ms}$.
[Assume, $\uparrow$ for positive $y-$ direction and $\downarrow$ for negative $y-$ direction]

• A. $5\text{cm}\uparrow$
• B. $5\text{cm}\downarrow$
• C. $1\text{cm}\uparrow$
• D. $1\text{cm}\downarrow$

Answer 1

The correct option is D $1\text{cm}\downarrow$

According to the principle of superposition of two waves,
$\overrightarrow{y_{net}}=\overrightarrow{y_1}+\overrightarrow{y_2}$
At $x=4.5\text{cm}$ and $t=5\text{ms}=5\times {10}^{-3}\text{s}$

${\overrightarrow{y}}_1=\left(2\text{cm}\right)\sin \left[\right(2\pi \times 4.5)-(100\pi \times 5\times {10}^{-3}\left)\right]$
$=\left(2\text{cm}\right)\sin [9\pi -0.5\pi ]=2\text{cm}\uparrow$
and
${\overrightarrow{y}}_2=\left(3\text{cm}\right)\sin \left[\right(\pi \times 4.5)-(200\pi \times 5\times {10}^{-3}\left)\right]$
$=\left(3\text{cm}\right)\sin [4.5\pi -\pi ]=3\text{cm}\downarrow$
So, we have
${\overrightarrow{y}}_{net}={\overrightarrow{y}}_1+{\overrightarrow{y}}_2=2-3=1\text{cm}\downarrow$

Hence, option (d) is the correct answer.