Question

Column I shows four systems, each of the same length $L$, for producing standing waves. The lowest possible natural frequency of a system is called its fundamental frequency, whose wavelength is denoted as ${\lambda }_f$. Match each system with statements given in Column II describing the nature and wavelength of the standing waves:

Column I		Column II
A	Pipe closed at one end	P	Longitudinal waves
B	Pipe open at both ends	Q	Transverse waves
C	Stretched wire clamped at both sides	R	${\lambda }_f=L$
D	Stretched wire clamped at both ends and at mid-point	S	${\lambda }_f=2L$
		T	${\lambda }_f=4L$

• A. $(A)\to (P),(T);(B)\to (P),(S);(C)\to (Q),(S);(D)\to (Q)(R)$
• B. $(A)\to (P),(R);(B)\to (P),(S);(C)\to (Q),(S);(D)\to (Q),(S)$
• C. $(A)\to (P),(S);(B)\to (P),(S);(C)\to (Q),(S);(D)\to (P)(R)$
• D. $(A)\to (P),(T);(B)\to (P),(S);(C)\to (R),(S);(D)\to (Q),(S)$

Answer 1

Answer: (A)

(A) $\to (P),(T)$
For a pipe closed at one end, we have
$\frac{{\lambda }_f}{4}=L\text{ or }{\lambda }_f=4L$
Therefore, the sound waves are longitudinal.
(B) $\to$ (P), (S)
For a pipe open at both ends, we have
$\frac{{\lambda }_f}{2}=L\text{ or }{\lambda }_f=2L$
Therefore, the sound waves are longitudinal.
(C) $\to \text{ (Q), (S) }$
For a stretched wire clamped at both ends, we have
${\lambda }_f/2=L$
Vibration on the string is transverse.
(D) $\to$ (Q), (R)
For a stretched wire clamped at both ends and at mid-point, we have
$\frac{{\lambda }_f}{2}=\frac{L}{2}\Rightarrow {\lambda }_f=L$
Therefore, the vibration on the string is transverse.