Question

A stretched wire fixed at both ends is oscillating in the third overtone mode. Equation of the transverse stationary wave produced in this wire is $y=A\sin \left(6\pi x\right)\sin \left(20\pi t\right)$ ($x$ is in $\text{m}$ and $t$ in $\text{seconds}$). Find the length of the wire.

• A. $\frac{2}{3}\text{m}$
• B. $\frac{3}{2}\text{m}$
• C. $\frac{4}{3}\text{m}$
• D. $\frac{3}{4}\text{m}$

Answer 1

The correct option is A $\frac{2}{3}\text{m}$

We have the wave equation for transverse stationary wave as.

$y=A\sin \left(kx\right)\cos \left(\omega t\right)$

From the given equation, we can see that
$k=6\pi {\text{m}}^{-1}$

$\therefore \lambda =\frac{2\pi }{k}=\frac{2\pi }{6\pi }=\frac{1}{3}\text{m}$

Now, in the third overtone mode, four loops are present as shown below.


The size of one loop is $\frac{\lambda }{2}$, therefore total length of the wire is,

$l=4\times \frac{\lambda }{2}=2\lambda =2\times \frac{1}{3}$

$\Rightarrow l=\frac{2}{3}\text{m}$