SATHEE: Wave Motion 2 Question 11

Wave Motion 2 Question 11

11. A pipe of length $85 c m$ is closed from one end. Find the number of possible natural oscillations of air column in the pipe whose frequencies lie below $1250 H z$ . The velocity of sound in air is $340 m / s$ .

(2014 Main)

(a) 12

(b) 8

(d) 4

Show Answer

Answer:

Correct Answer: 11. (c)

Solution:

For closed organ pipe $= \frac{(2 n + 1) v}{4 l} [n = 0, 1, 2 \dots \dots$ .

$\begin{array}{r} \frac{(2 n + 1) v}{4 l} < 1250 \\ (2 n + 1) < 1250 \times \frac{4 \times 0.85}{340} \\ (2 n + 1) < 12.52 n < 11.50 \\ n < 5.25 \end{array}$

So,

So, we have 6 possibilities.

Alternate method

In closed organ pipe, fundamental node

i.e. $\frac{λ}{4} = 0.85 \Rightarrow λ = 4 \times 0.85$

As we know, $v = \frac{c}{λ} \Rightarrow \frac{340}{4 \times 0.85} = 100 H z$

$∴$ Possible frequencies $= 100 H z, 300 H z, 500 H z, 700 H z$ , $900 H z, 1100 H z$ below $1250 H z$ .

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Wave Motion 2 Question 11

11. A pipe of length 85cm is closed from one end. Find the number of possible natural oscillations of air column in the pipe whose frequencies lie below 1250Hz. The velocity of sound in air is 340m/s.

Answer:

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11. A pipe of length $85 c m$ is closed from one end. Find the number of possible natural oscillations of air column in the pipe whose frequencies lie below $1250 H z$ . The velocity of sound in air is $340 m / s$ .