Question

A string, fixed at both ends, vibrates in a resonant mode with a separation of $2.0\text{cm}$ between the consecutive nodes. For the next higher resonant frequency, this separation is reduced to $1.6\text{cm}$. Find the length of the string.

• A. $6\text{cm}$
• B. $8\text{cm}$
• C. $10\text{cm}$
• D. $12\text{cm}$

Answer 1

The correct option is B $8\text{cm}$

Case I: When separation between the consecutive nodes is $2\text{cm}$.

Let the wavelength be ${\lambda }_1$

Thus, from the data given in the question we can say that,

$\frac{{\lambda }_1}{2}=2\text{cm}$

$\therefore {\lambda }_1=4\text{cm}$

Let there be $n$ loops in this case
$\Rightarrow$ length of the wire, $l=\frac{\left(n{\lambda }_1\right)}{2}\dots \left(i\right)$

Case II:When separation between the consecutive nodes is $1.6\text{cm}$.

Let the wavelength be ${\lambda }_2$
Thus, from the data given in the question we can say that,

$\frac{{\lambda }_2}{2}=1.6\text{cm}$

$\Rightarrow {\lambda }_2=3.2\text{cm}$

Since, the string is fixed at both ends, the next higher frequency will have $(n+1)$ loops.

So there are $(n+1)$ loops with the II case

$\therefore$ length of the wire, $l=\frac{(n+1){\lambda }_2}{2}.....\dots \left(ii\right)$

From equation ($i$) and ($ii$)

we can write that, $\frac{n{\lambda }_1}{2}=\frac{(n+1){\lambda }_2}{2}$
$\Rightarrow n\times 2=(n+1)\times \left(1.6\right)$
$\Rightarrow n=4$

Substitute equation $n$ in equation ($i$)
$\therefore$ length of the string, $l=\frac{\left(n{\lambda }_1\right)}{2}$
$\Rightarrow l=4\times 2=8\text{cm}$

Hence, option (b) is the correct answer.