Question

The transverse displacement of a string is given by $y(x,t)=0.06\sin \left(\frac{2\pi }{3}x\right)\cos \left(120\pi t\right)$ where $x$ and $y$ are in $m$ and $t$ in $\text{sec}$. It is given that the string vibrates in its third harmonic. Which among the following is/are the position(s) of nodes?

• A. $1\text{m}$
• B. $1.5\text{m}$
• C. $3\text{m}$
• D. $4\text{m}$

Answer 1

Answer: (B), (C)

The correct options are
B $1.5\text{m}$
C $3\text{m}$

Given that,
$y(x,t)=0.06\sin \left(\frac{2\pi }{3}x\right)\cos \left(120\pi t\right)$
Compare above equation with
$y(x,t)=A\sin kx\cos \omega t$
$k=\frac{2\pi }{\lambda }=\frac{2\pi }{3}$
$\Rightarrow \lambda =3\text{m}$


$\Rightarrow \frac{\lambda }{2}=1.5\text{m}$
Hence, positions of nodes are $x=0,1.5\text{m},3\text{m}$ and $4.5\text{m}$
Options (b) and (c) are correct.