Question 20: Waves

Question

The number of possible natural oscillations of air column in a pipe closed at one end of length $$85 \mathrm{~cm}$$ whose frequencies lie below $$1250 \mathrm{~Hz}$$ are (velocity of sound $$=340 \mathrm{~ms}^{-1}$$)

Accepted Answer

For a pipe closed at one end, the fundamental frequency is given by:
$$f_1 = \frac{v}{4L}$$
where:
$$f_1$$ is the fundamental frequency
$$v$$ is the speed of sound
$$L$$ is the length of the pipe
The possible frequencies for a pipe closed at one end are odd multiples of the fundamental frequency:
$$f_n = (2n-1)f_1 = (2n-1)\frac{v}{4L}$$
where $$n$$ is an integer (1, 2, 3, ...).
We need to find the number of possible frequencies below $$1250 \mathrm{~Hz}$$.  Let's plug in the given values and solve for $$n$$:
$$1250 \mathrm{~Hz} \ge (2n-1)\frac{340 \mathrm{~ms}^{-1}}{4(0.85 \mathrm{~m})}$$
Simplifying the inequality:
$$1250 \mathrm{~Hz} \ge (2n-1) 100 \mathrm{~Hz}$$
$$12.5 \ge 2n - 1$$
$$13.5 \ge 2n$$
$$n \le 6.75$$
Since $$n$$ must be an integer, the maximum value of $$n$$ is 6. This means there are
6
possible frequencies below $$1250 \mathrm{~Hz}$$.
Therefore, the correct answer is
Option A: 6
.

Answer

6

Answer

8

Answer

4

Answer

5

Waves

The number of possible natural oscillations of air column in a pipe closed at one end of length $85 \mathrm{~cm} whose frequencies lie below 1250 \mathrm{~Hz} are (velocity of sound =340 \mathrm{~ms}^{-1}$)

Choose the correct answer: