wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

An open organ pipe has a length of 10 cm . If the speed of sound in air is 340 m/s and the audible range is (20−20,000 Hz), then choose the correct option(s):

A
The fundamental frequency of vibration of this pipe is 1700 Hz.
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
11 is the highest harmonic of such tube in audible range.
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
6800 Hz is the third overtone frequency.
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
The fundamental frequency of vibration of this pipe is 1200 Hz.
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C 6800 Hz is the third overtone frequency.
Given that,
Length of pipe, L=10 cm or 0.1 m
Speed of sound, v=340 m/s

Mode of vibration of air column in an open organ pipe is given by

fn=nv2L where n=1,2,3...

Fundamental frequency, f1=v2L
From the data given in the question,
f1=3402×10×102=1700 Hz

For maximum audible frequency,
fn=20000 Hz
So
fn=nv2L
n=20000×2×0.1340=11.7611

For third overtone or fourth harmonic,

f4=4v2L=4×3402×0.1=6800 Hz

Thus, options (a) , (b) and (c) are the correct answers.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Standing Longitudinal Waves
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon