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 $x=0.10\text{m}$. The displacement $y\left(t\right)$ of a particle on the string particle at $x=0$ is shown in figure. Then,

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

Answer 1

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

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

From graph,
$\frac{\lambda }{4}=0.1\Rightarrow \lambda =0.4m$
from graph $\Rightarrow T=0.2\text{sec}$. and amplitude of standing wave is $2A=4\text{cm}$.
At $x=0,$ there is antinode
Amplitude $A^{\prime }=2A\cos kx$
At $t=0,y=0,$ hence other part of stationary wave equation is $\sin \omega t.$

Equation of the standing wave
$y(x,t)=-2A\cos \left(\frac{2\pi }{0.4}x\right).\sin \left(\frac{2\pi }{0.2}t\right)\text{cm}$
speed $=\frac{\lambda }{T}=2\text{m/sec}$
$y(x=0.05,t=0.05)=-2\sqrt{2}\text{cm}$
$y(x=0.04,t=0.05)=-4\cos {36}^{\circ }$
$V_y=\frac{dy}{dt}=-2A\times \frac{2\pi }{0.2}\cos \left(\frac{2\pi x}{0.4}\right).\cos \left(\frac{2\pi t}{0.2}\right)$
$V_y{|}_{(x=\frac{1}{15}m,t=0.1s)}=20\pi \text{cm/sec}$

Why this question? Caution: The x-axis does not have x-coordinate, so calculation of wavelength from graph could go wrong.