CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

An air column in a tube 32 cm long, closed at one end, is in resonance with a tuning fork. The air column in another tube, open at both ends, of length 66 cm is in resonance with another tuning fork. When these two tuning forks are sounded together, they produce 8 beats per second. Then the frequencies of the two tuning forks are:
(Consider fundamental frequencies only)

A
250 Hz,258 Hz
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
240 Hz,248 Hz
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
264 Hz,256 Hz
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
280 Hz,272 Hz
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C 264 Hz,256 Hz
Since v=nλ and velocity in two cases must be same as the medium is same. λ1=4(32) cm=128 cm and λ2=2(66) cm=132 cm.

Hence, n1n2=λ2λ1=132128=3332.

Also, n1n2=8,

Solving these gives n1=264Hz and n1=256Hz.

flag
Suggest Corrections
thumbs-up
3
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