Question

When a stretched wire and a tuning fork are sounded together $5$ beats per second are heard in both cases, when the lengths of stretched wire were $100cm$ or $95cm$ in fundamental mode with same tension. Then frequency of tuning fork is $(Hz)$

• A. 105
• B. 100
• C. 175
• D. 195

Answer 1

Answer: (D)

$n_0=\frac{1}{2l}\sqrt{\frac{T}{\mu }}$
Let frequency of fork is $n$
$n_1=\frac{1}{2l_1}\sqrt{\frac{T}{\mu }}$
$n_2=\frac{1}{2l_2}\sqrt{\frac{T}{\mu }}$
dividing equations we get
$\frac{n_1}{n_2}=\frac{l_2}{l_1}=\frac{95}{100}$
$n_2-n_1=10$
$n_2-\frac{19}{20}{n}_2=10$
$\frac{20n_2-19n_2}{20}=10$
$n_2=200$
$\therefore n=200-5$
$=195Hz$