First slide

Beats

Question

The frequency of tuning forks A and B are respectively 3% more and 2% less than the frequency of tuning fork C. When A and B are simultaneously excited, 5 beats per second are produced. Then the frequency of the tuning fork 'A' (in Hz) is

Moderate

A

98
B

100
C

103
D

105

Solution

Let n be the frequency of fork C then
$n_{A} = n + \frac{3 n}{100} = \frac{103 n}{100} and n_{B} = n - \frac{2 n}{100} = \frac{98}{100}$
but $n_{A} - n_{B} = 5 \Rightarrow \frac{5 n}{100} = 5 \Rightarrow n = 100 Hz$
$∴ n_{A} = \frac{(103) (100)}{100} = 103 Hz$

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