Question

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

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

Answer 1

Answer: (C)

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