Question

Two tuning forks A and B give 6 beats per second. A resonates with a closed column of air 15 cm long and B with an open column 30.5 cm long in their fundamental harmonics. Their frequencies are

• A. n${}_A$
  = 366Hz, n${}_B$ = 360Hz
• B. n${}_A$
  = 466Hz, n${}_B$ = 460Hz
• C. n${}_A$
  = 566Hz, n${}_B$ = 560Hz
• D. n${}_A$
  = 766Hz, n${}_B$ = 760Hz

Answer 1

The correct option is A n${}_A$

= 366Hz, n${}_B$ = 360Hz

$f_A=\frac{\vartheta }{4\left(15\right)}[\therefore \text{fundamental frequency in closed is}\frac{\vartheta }{4l}]$

$f_B=\frac{\vartheta }{2\left(30.5\right)}[\therefore \text{fundamental frequency in open is}\frac{\vartheta }{2l}]$

$beat=f_A-f_B$

$6=\frac{\vartheta }{4\left(15\right)}-\frac{\vartheta }{2\left(30.5\right)}$

$\vartheta =\frac{6\times 2\times \left(30.5\right)\times 30}{30.5-30}$$=21960cm/s$

$f_A=\frac{21960}{4\left(15\right)}=366Hz$

$f_B=\frac{21960}{2\left(30.5\right)}=360Hz$