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 m. The displacement y(t) of a particle on the string particle at x = 0 is shown in figure. Then:

• A. $\lambda =0.4m$
• B. Speed of the particle at $x=\frac{1}{15}m\text{and}t=0.1s$ is $10\pi cm/sec$
• C. amplitude of standing wave is $=4\text{cm}$
• D. wave speed is 2 m/s

Answer 1

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

wave speed is 2 m/s

$\frac{\lambda }{4}=0.1\Rightarrow \lambda =0.4m$
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 $=\frac{\lambda }{T}=2\text{m/sec}$
$V_y=\frac{dy}{dt}=-2A\times \frac{2\pi }{0.2}\cos \left(\frac{2\pi x}{0.4}\right).\cos \left(\frac{2\pi t}{0.2}\right)$
$V_y{|}_{(x=\frac{1}{15}m,t=0.1s)}=20\pi \text{cm/sec}$

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