Question

For a certain transverse standing wave on a long string, an antinode is formed at $x=0$ and next to it, a node is formed at $x=0.10m$. The position $y\left(t\right)$ of the string particle at $x=0$ is shown in figure.

• A. Transverse displacement of the particle at $x=0.05m$ and $t=0.05s$ is $-2\sqrt{2}cm$.
• B. Transverse displacement of the particle at $x=0.04m$ and $t=0.025s$ is $-2\sqrt{2}cm$.
• C. Speed of the travelling waves that interfere to produce this standing wave is $2m/s$.
• D. The transverse velocity of the string particle at $x=\frac{1}{15}m$ and $t=0.1s$ is $20\pi cm/s$.

Answer 1

Answer: (A), (C), (D)

The transverse velocity of the string particle at $x=\frac{1}{15}m$ and $t=0.1s$ is $20\pi cm/s$.

Distance between an antinode and a node is equal to one fourth of the wavelength of the wave.
$\therefore \frac{\lambda }{4}=0.1\Rightarrow \lambda =0.4m$
From graph $\Rightarrow T=0.2$ s
and amplitude of standing wave is $2A=4cm.$
Equation of the standing wave
$\begin{array}{cc}y(x,t)& =-2A\cos \left(\frac{2\pi }{0.4}x\right)\sin \left(\frac{2\pi }{0.2}t\right)cm\\ y(x=0.05,& t=0.05)=-2\sqrt{2}cm\end{array}$
$y(x=0.04,t=0.025)=-2\sqrt{2}\cos {36}^{\circ }$
Speed $=\frac{\lambda }{T}=2m/s$
$$\begin{array}{cc}& V_y=\frac{dy}{dt}=-2A\times \frac{2\pi }{0.2}\cos \left(\frac{2\pi x}{0.4}\right)\cos \left(\frac{2\pi t}{0.2}\right)\\ & V_y=(x=\frac{1}{15}m,t=0.1)=20\pi cm/s\end{array}$$