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 $100\, cm$ or $95\, cm$ 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)

$n_{0}=\frac{1}{2 l} \sqrt{\frac{T}{\mu}}$
Let frequency of fork is $n$
$n_{1}=\frac{1}{2 l_{1}} \sqrt{\frac{T}{\mu}}$
$n_{2}=\frac{1}{2 l_{2}} \sqrt{\frac{T}{\mu}}$
dividing equations we get
$\frac{n_{1}}{n_{2}}=\frac{l_{2}}{l_{1}}=\frac{95}{100}$
$n_{2}-n_{1}=10$
$n_{2}-\frac{19}{20} n_{2}=10$
$\frac{20 n_{2}-19 n_{2}}{20}=10$
$n_{2}=200$
$\therefore n=200-5$
$=195\, Hz$