JEE MAIN - Physics (2014 (Offline) - No. 17)
A pipe of length $$85$$ $$cm$$ is closed from one end. Find the number of possible natural oscillations of air column in the pipe whose frequencies lie below $$1250$$ $$Hz$$. The velocity of sound in air is $$340$$ $$m/s$$.
$$12$$
$$8$$
$$6$$
$$4$$
Explanation
Length of pipe $$=85$$ $$cm$$ $$=0.85m$$
Pipe is closed from one end so it behaves as a closed organ pipe
Frequency of oscillations of air column in closed organ pipe is given by,
$$f = {{\left( {2n - 1} \right)\upsilon } \over {4L}}$$
$$f = {{\left( {2n - 1} \right)\upsilon } \over {4L}} \le 1250$$
$$ \Rightarrow {{\left( {2n - 1} \right) \times 340} \over {0.85 \times 4}} \le 1250$$
$$ \Rightarrow 2n - 1 \le 12.5 \approx 6$$
Possible value of n = 1, 2, 3, 4, 5, 6
So, number of possible natural frequencies lie below 1250 Hz is 6.
Pipe is closed from one end so it behaves as a closed organ pipe
Frequency of oscillations of air column in closed organ pipe is given by,
$$f = {{\left( {2n - 1} \right)\upsilon } \over {4L}}$$
$$f = {{\left( {2n - 1} \right)\upsilon } \over {4L}} \le 1250$$
$$ \Rightarrow {{\left( {2n - 1} \right) \times 340} \over {0.85 \times 4}} \le 1250$$
$$ \Rightarrow 2n - 1 \le 12.5 \approx 6$$
Possible value of n = 1, 2, 3, 4, 5, 6
So, number of possible natural frequencies lie below 1250 Hz is 6.
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