Question

The equation of a stationary wave (all quantities are in SI units) y = (0.06) sin $\left(2\pi x\right)$ cos $\left(5\pi t\right)$
$\begin{array}{cccc}& Column-I& & Column-II\\ \left(a\right)& \text{Amplitude of constituent wave}& \left(p\right)& \text{0.06}\\ \left(b\right)& \text{Position of node is at}{x}_{1}equalto& \left(q\right)& \text{0.5}\\ \left(c\right)& \text{Position of anti node is at}{x}_2\text{equal to}& \left(r\right)& \text{0.25}\\ \left(d\right)& \text{Amplitude at}x=\frac{3}{4}mis& \left(s\right)& \text{0.03}\end{array}$

• A. (a) $\to$ (s); (b) $\to$ (q); (c) $\to$ (r); (d) $\to$ (p);
• B. (a) $\to$ (p); (b) $\to$ (q); (c) $\to$ (r); (d) $\to$ (s);
• C. (a) $\to$ (s); (b) $\to$ (r); (c) $\to$ (q); (d) $\to$ (p);
• D. (a) $\to$ (p); (b) $\to$ (r); (c) $\to$ (q); (d) $\to$ (s);

Answer 1

The correct option is A (a) $\to$ (s); (b) $\to$ (q); (c) $\to$ (r); (d) $\to$ (p);

y = 2A sin kx cos $\omega t$

$\therefore$ 2A = 0.06 $\Rightarrow$ A = 0.03 m

$\to$ At nodes y = 0 $\Rightarrow$ sin $2\pi x=0\Rightarrow x=0.5$ m

$\to$ At antinodes y = maximum $\Rightarrow$ sin $2\pi x=1$

$\Rightarrow x=0.25$m

$\to |y_{max}|=|2Asin2\pi \times \frac{3}{4}|=0.06$m