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

A tuning fork produces 8 beats per second when sounded together with a fork of frequency 340 Hz. When the prongs of tuning fork C are filled a little, the number of beats produced per second decreases to 4. Find the frequency of the tuning fork C before filling its prongs.

A
332 Hz
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
434 Hz
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
430 Hz
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
436 Hz
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 332 Hz
given :

nD= tuning fork D frequency =340Hz
nC= tuning fork C frequency=84=4 beats per second.
First nC±nD=8(before filing)
nC±nD=4(after filing)
From given condition nc±nD=8
nc±340=8
nc=340+8=348Hz
or,nc=3408=332Hz
when tuning fork C is filed then nc±nD=4
nc±340=4
nc=340+4=344Hz
or,nc=3404=336Hz
The frequency of tuning fork increases on filing. Hence nc344Hz.
If original frequency of tuning fork C is taken 332 Hz, then on filing both the value 344Hz, and 336Hz are greater.
Also it produces 4 beats per second with tuning fork D.
frequency of tuning fork C=332Hz
Hence,
option (A) is correct answer.

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