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

A tuning fork produces 4 beats per second with another tuning fork of frequency 256 Hz. The first one is now loaded with a little wax and the beat frequency is found to increase to 6 per second. What was the original frequency of the tuning fork?

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

The correct option is A 252 Hz

A tuning fork produces 4 beats with a known tuning fork whose frequency = 256 Hz

So the frequency of unknown tuning fork = either 256 - 4 = 252 or 256 + 4 = 260 Hz

Now as the first one is loaded with wax its mass/unit length increases. So velocity will decrease.

So, its frequency decreases. As frequency decreases the beat frequency increases ( becomes 6 from 4)

So original frequency must be 252 Hz.

260 Hz is not possible as on decreasing the frequency the beats should decrease which is not happening here.


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