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The string fixed at both ends has standing wave nodes for which distance between adjacent nodes is $x_{1}$ . The same string has another standing wave nodes for which distance between adjacent nodes is $x_{2}$ . If l is the length of the string, then $x_{2} / x_{1} = l / (l + 2 x_{1})$ . What is the difference in numbers of the loops in the two cases?

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answer is B.

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Detailed Solution

Let no. of loops formed in first case $= n$

$x_{1} n = l ....... (1)$

Let no. of loops formed in second case

$= (n + k)$

$x_{2} (n + k) = l ..... (2)$

From (1) and (2), $x_{2} [\frac{l}{x_{1}} + k] = l$

$\frac{x_{2}}{x_{1}} = \frac{l}{l + k x_{1}}$

Comparing with $\frac{x_{2}}{x_{1}} = \frac{l}{l + 2 x_{1}}, k = 2$

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