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 correct options are
A Transverse displacement of the particle at $x=0.05m$ and $t=0.05s$ 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 $20\pi cm/s.$

Distance between an antinode and a node is equal to one fourth of the wavelength $\lambda$ of the wave.

Thus $\frac{\lambda }{4}=0.1\Rightarrow \lambda =0.4m$

From the graph, we get time period of wave $T=0.2sec$ and amplitude of the standing wave is $2A=4cm.$

Equation of the standing wave

$y(x,t)=-2A\cos \left(\frac{2\pi }{0.4}x\right)\sin \left(\frac{2\pi }{0.2}t\right)cm$

Putting the values in the above equation, we get

$y(x=0.05,t=0.05)$ $=-\left(4\right)\cos \left(\frac{2\pi }{0.4}\right(0.05\left)\right)\sin \left(\frac{2\pi }{0.2}\right(0.05\left)\right)cm$

$y(x=0.05,t=0.05)$ $=-\left(4\right)\cos \left(\frac{\pi }{4}\right)\sin \left(\frac{\pi }{2}\right)cm$ $=-2\sqrt{2}cm$

$y(x=0.04,t=0.025)$ $=-\left(4\right)\cos \left(\frac{2\pi }{0.4}\right(0.04\left)\right)\sin \left(\frac{2\pi }{0.2}\right(0.025\left)\right)cm$

$y(x=0.04,t=0.025)$ $=-\left(4\right)\cos \left(\frac{\pi }{5}\right)\sin \left(\frac{\pi }{4}\right)cm$ $=-2\sqrt{2}\cos {36}^{o}cm$

Speed $v=\frac{\lambda }{T}=\frac{0.4}{0.2}=2m/s$

Transverse velocity of wave $V_y=\frac{dy}{dt}=-2A\times \frac{2\pi }{0.2}cos\left(\frac{2\pi x}{0.4}\right)cos\left(\frac{2\pi t}{0.2}\right)$

$V_y(x=\frac{1}{15}cm,t=0.1s)=\frac{dy}{dt}=-\left(4\right)\times \frac{2\pi }{0.2}cos\left(\frac{2\pi }{0.4\left(15\right)}\right)cos\left(\frac{2\pi \left(0.1\right)}{0.2}\right)$

$V_y(x=\frac{1}{15}cm,t=0.1s)=\frac{dy}{dt}=-\left(4\right)\times \frac{2\pi }{0.2}cos\left(\frac{\pi }{3}\right)cos\left(\pi \right)$

Thus we get $V_y(x=\frac{1}{15}cm,t=0.1s)=-40\pi \times 0.5\times -1=20\pi cm/s$