Question

Spacing between two successive nodes in a standing wave on a string is $x$. If frequency of the standing wave is kept unchanged but tension in the string is doubled, then new spacing between successive nodes will become:

• A. $2x$
• B. $\sqrt{2}x$
• C. $\frac{x}{2}$
• D. $\frac{x}{\sqrt{2}}$

Answer 1

The correct option is B $\sqrt{2}x$

Spacing between successive nodes is $x$.
$\Rightarrow x=\frac{{\lambda }_1}{2}$
$\Rightarrow {\lambda }_1=2x$
Now, $v=f\lambda$
Here, frequency is kept fixed, Hence
$\lambda \propto v$
$\lambda \alpha \sqrt{\frac{T}{\mu }}$
where T is tension in the string and $\mu$is linear mass density

$\lambda \alpha \sqrt{\frac{T}{\mu }}$
$\frac{{\lambda }_2}{{\lambda }_1}=\frac{\sqrt{T_2}}{\sqrt{T_1}}$
($\because$ $\mu$ is same for both strings)
$\frac{{\lambda }_2}{{\lambda }_1}=\frac{\sqrt{2T}}{\sqrt{T}}$
${\lambda }_2=\sqrt{2}{\lambda }_1$
New spacing between nodes is
$x\text{'}=\frac{{\lambda }_2}{2}=\frac{\sqrt{2}{\lambda }_1}{2}$
$\Rightarrow x\text{'}=\frac{{\lambda }_1}{\sqrt{2}}$
From equation $\left(i\right)$
$x\text{'}=\frac{2x}{\sqrt{2}}=\sqrt{2}x$
Final Answer: $\left(b\right)$