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

The harmonic mode which resonates with a closed pipe of length 22cm, when excited by a 1875Hz source and the number of nodes present in it respectively are (velocity of sound in air =330ms1):

A
1st,1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
3rd,1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
3rd,2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
5th,4
No worries! We‘ve got your back. Try BYJU‘S free classes today!
E
5th,3
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is E 5th,3
Velocity of sound=330ms1
Length of closed pipe=22cm=22×102m
Fundamental frequency of stationary wave,
v0=v4l=3304×22×102
v0=330×1024×22=30008
v0=375Hz
3v0=375×3=1125
5v0=5×375=1875
Thus we get 5th harmonic
and number of nodes =3
Note: In closed pipe, odd harmonics are produced.

flag
Suggest Corrections
thumbs-up
2
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Normal Modes on a String
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon