Question

For a certain transverse standing wave having wavelength $0.4\text{m}$ 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(t) of the string particle at $x=0$ is shown in figure.

• 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 $t=0.025\text{~}s$ 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 correct options are
A Transverse displacement of the particle at $x=0.05\text{m}$ 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 $20\pi \text{cm/s}$

We know that the general form of standing wave is
$S_x=A\cos kx\sin \omega t$

Wavelength is the distance covered in one time period
So $\lambda =0.4\text{m}$ (given)
$T=0.2\text{s}$ (from diagram)

(A) $S_x=-4\cos kx\sin \omega t$
$=-4\cos (\frac{2\pi }{0.40}\times 0.05)\sin (\frac{2\pi }{0.2}\times 0.05)=-2\sqrt{2}\text{cm}$

(C) $\frac{\lambda }{T}=v=\frac{0.4}{0.2}=2\text{m/s}$

(D) For standing wave, velocity is given by
$v_p=-A\omega \cos kx\sin \omega t$
$\omega =\frac{2\pi }{T}=10\pi s^{-1}$
$v_p=-4\times 10\pi \cos (\frac{2\pi }{0.4}\times \frac{1}{15})\times \frac{2\pi }{0.2}\times \cos (\frac{2\pi }{0.2}\times 0.1)$
$=+40\pi \times \frac{1}{2}=20\pi \text{cm/s}$