Two identical wires have the same fundamental frequency of $400 Hz$. when kept under the same tension. If the tension in one wire is increased by $2\%$ the number of beats produced will be
A$4$
B$2$
C$8$
D$1$
Medium
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A$4$
a (a) $n \propto \sqrt T $ ==> $\frac{{\Delta n}}{n} = \frac{1}{2}\frac{{\Delta T}}{T}$
Beat frequency $ = \Delta n = \left( {\frac{1}{2}\frac{{\Delta T}}{T}} \right)\,n = \frac{1}{2} \times \frac{2}{{100}} \times 400 = 4$
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