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	$\lambda_{f}=2 L$
		T	$\lambda_{f}=4 L$

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

Answer 1

Answer: (A)

(A) $\rightarrow( P ),( T )$
For a pipe closed at one end, we have
$\frac{\lambda_{f}}{4}= L \text { or } \lambda_{f}=4 L$
Therefore, the sound waves are longitudinal.
(B) $\rightarrow$ (P), (S)
For a pipe open at both ends, we have
$\frac{\lambda_{f}}{2}= L \text { or } \lambda_{f}=2 L$
Therefore, the sound waves are longitudinal.
(C) $ \rightarrow \text { (Q), (S) }$
For a stretched wire clamped at both ends, we have
$\lambda_{f} / 2= L$
Vibration on the string is transverse.
(D) $\rightarrow$ (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.