The tuning forks A and B produce notes of frequency 256 Hz and 262 Hz respectively. An unknown note sounded at the sametime as A produces beats. When the same note is sounded with B, beat frequency is twice as large. The unknown frequency could be:
A
268 Hz
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
250 Hz
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
260 Hz
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
None of these
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is B 250 Hz a is the +ve constant (I)256+a=262−2a=f0⇒a=2 f0=258Hz (II)256+a=262+2a=f0
Not Possible (III)256+a=262+2a=f0
Not possible a have -ve value (IV)256−a=262−2a=f0⇒a=6 f0=250Hz