8 beats/s are produced when two tuning forks A, B are made to vibrate together. The number of beats decreases to 5 beats/s when prongs of B are filed. If natural frequency of A is 282 Hz, the frequency of B before and after filing is

274 Hz, 277 Hz

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

290 Hz, 287 Hz

No worries! We‘ve got your back. Try BYJU‘S free classes today!

290 Hz, 277 Hz

No worries! We‘ve got your back. Try BYJU‘S free classes today!

274 Hz, 290 Hz

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A 274 Hz, 277 Hz
By filing frequency of B is increased and beats is decreasing, so, frequency of B is less than frequency of A.
$b e a t s = δ_{A} - δ_{B}$
$8 = 282 - δ_{B}$
$\Rightarrow δ_{B} = 274 H_{z}$
After filling beats become $5$ .
$b e a t s = δ_{A} - δ_{B}$
$5 = 282 - δ_{B}$
$\Rightarrow δ_{B} = 277 H_{z}$

Suggest Corrections

Similar questions

Q. A tuning fork

C

produces

8

beats per second with another tuning for

D

of frequency

340 H z

. When the prongs of tuning fork

C

are filed as little, then number of beats produced per second decreases to

4

. Find the frequency of the tuning fork

C

before filing its prongs.

When two tuning forks A and B are sounded together, 4 beats per second are heard. The frequency of the fork B is 384 Hz. When one of the prongs of the fork A is filed and sounded with B, the beat frequency increases, then the frequency of the fork A is

Q. A tuning fork produces