The number of possible natural oscillations of air column in a pipe closed at one end of length 85 cm whose frequencies lie below 1250 Hz are (velocity of sound =340ms−1 )
For a pipe closed at one end, the fundamental frequency is given by: f1=‌
v
4L
where: f1 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: fn=(2n−1)f1=(2n−1)‌
v
4L
where n is an integer (1,2,3,...). We need to find the number of possible frequencies below 1250 Hz . Let's plug in the given values and solve for n : 1250‌Hz≥(2n−1)‌
340ms−1
4(0.85m)
Simplifying the inequality: ‌1250‌Hz≥(2n−1)100‌Hz ‌12.5≥2n−1 ‌13.5≥2n ‌n≤6.75 Since n must be an integer, the maximum value of n is 6 . This means there are 6 possible frequencies below 1250 Hz . Therefore, the correct answer is Option A: 6.