Question

A string fixed at both ends has a standing wave mode for which the distances between adjacent nodes is $18 \, cm.$ For the net consecutive standing wave mode distance between adjacent nodes is $16 \, cm.$ The minimum possible length of the string is

• A. $144 \, \, \text{cm}$
• B. $204 \, \, \text{cm}$
• C. $288 \, \, \text{cm}$
• D. $72 \, \, \text{cm}$

Answer 1

Answer: (A)

Let the vibration takes place in the nth mode
So, for $1^{s t}$ case, $\frac{n \lambda }{2}=L$ ......(i)
And for $2^{\text {nd }}$ case, $(n+1) \frac{\lambda^{\prime}}{2}=L \ldots $..... (ii)
From Equations. (i) and (ii), we get
$n\frac{\lambda }{2}=\left(\right.n+1\left.\right)\frac{\lambda ′}{2}$ $\left[\because \, \, \, \frac{\lambda }{2} = 18 cm \, a n d \frac{\lambda ^{′}}{2} = 16 cm\right]$
$\Rightarrow \, \, \, 18n=\left(n + 1\right)16$
$\Rightarrow \, n=8$
So, minimum possible length $l=\frac{n \lambda }{2}$
$\Rightarrow \, l=8\times 18=144 \, cm$