Question

In a closed tube when air column is 20 cm it is in resonance with tuning fork A. When the length is increased by 2 cm then the air column is in resonance with tuning fork B. When A and B are sounded together they produce 8 beats per second. The frequencies of the tuning forks A and B are (in Hz) :

• A. 40, 44
• B. 88, 80
• C. 80, 88
• D. 44, 40

Answer 1

Answer: (B)

88, 80

Initial length of closed tube $L_i=20cm=0.2m$
Frequency of tuning fork A ${\nu }_A=\frac{v}{4L_i}=\frac{v}{4\left(0.2\right)}$
New length of the tube $L_f=22cm=0.22m$
Frequency of tuning fork B ${\nu }_B=\frac{v}{4L_f}=\frac{v}{4\left(0.22\right)}$
Beat frequency $b={\nu }_A-{\nu }_B$
$\therefore$ $8=\frac{v}{4\left(0.2\right)}-\frac{v}{4\left(0.22\right)}$
$⟹$ $v=70.4m/s$
Frequency of tuning fork A ${\nu }_A=\frac{70.4}{4\left(0.2\right)}=88Hz$
Frequency of tuning fork B ${\nu }_B=\frac{70.4}{4\left(0.22\right)}=80Hz$