Column I shows four systems, each of the same length L, for produ… - Vidyadip

MCQ

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.

A
$(A) \rightarrow p, r, \ and \ t,(B) \rightarrow p \ and \ q (C) \rightarrow q, \ and \ s,(D) \rightarrow r \ and \ s$
B
$(A) \rightarrow q \ and \ t,(B) \rightarrow r \ and \ s (C) \rightarrow p, \ and \ s,(D) \rightarrow q \ and \ t$
✓
$(A) \rightarrow p \ and \ t,(B) \rightarrow p \ and \ s (C) \rightarrow q, \ and \ s,(D) \rightarrow q \ and \ r$
D
$(A) \rightarrow q \ and \ t,(B) \rightarrow r \ and \ t (C) \rightarrow q \ and \ s,(D) \rightarrow s \ and \ t$

✓

Answer

Correct option: C.

$(A) \rightarrow p \ and \ t,(B) \rightarrow p \ and \ s (C) \rightarrow q, \ and \ s,(D) \rightarrow q \ and \ r$

c
In case of pipe we can produce longitudinal waves and for strings we produce transverse waves.

$A \rightarrow(p, t):$ It is a closed organ pipe hence in case of fundamental vibrations length of the tube must be $\left(\frac{1}{4}\right)^{\text {th }}$ of the wavelength (half loop).

$\frac{\lambda_t}{4}=L \Rightarrow \lambda_t=4 L$

$B \rightarrow(p, s)$ : It is open organ pipe hence in case of fundamental vibrations length of the pipe must be half of the wavelength (one loop).

$\frac{\lambda_t}{2}=L \Rightarrow \lambda_t=2 L$

$C \rightarrow(q, s)$ : It is a case of string of length $L$ fixed at both ends.

In case of fundamental vibrations length of the pipe must be half of the wavelength (one loop).

$\frac{\lambda_t}{2}=L \Rightarrow \lambda_t=2 L$

$D \rightarrow(q, r):$ In this case mid-point is fixed hence half-length must form one loop.

$\frac{\lambda_t}{2}=\frac{L}{2} \Rightarrow \lambda_t=L$

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