Question

The string fixed at both ends has standing wave nodes for which distance between adjacent nodes is $x_1$. The same string has another standing wave nodes for which distance between adjacent nodes is $x_2$. If $l$ is the length of the string, then $x_2=\frac{l}{(l+2x_1)}$. What is the difference in numbers of the loops in the two cases?

• A. $4$
• B. $2$
• C. $3$
• D. $1$

Answer 1

The correct option is B $2$

Let number of loops formed in first case $=n$
$x_{1}n=l\dots \left(i\right)$
Let number of loops formed in second case $=(n+k)$
$x_2(n+k)=l\dots \left(ii\right)$
From $\left(i\right)$ and $\left(ii\right)$,
$x_2[\frac{l}{x_1}+k]=l,\frac{x_2}{x_1}=\frac{l}{(l+k{x}_1)}$
Comparing with
$\frac{x_2}{x_1}=\frac{l}{(l+2x_1)},k=2$

Final answer: (b)