First slide

Standing waves

Question

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 $\frac{x_{2}}{x_{1}} = l (l + 2 x_{1})$ . What is the difference in numbers of the loops in the two cases?

Moderate

A

1
B

2
C

3
D

4

Solution

Let no. of loops formed in first case = n
$x_{1} n = l$ -------(i)
Let no. of loops formed in second case = (n + k)
$x_{2} (n + k) = l$ -----(ii)
From (i) and (ii) $x_{2} [\frac{l}{x_{1}} + k] = l, \frac{x_{2}}{x_{1}} = \frac{l}{l + {kx}_{1}}$
Comparing with $\frac{x_{2}}{x_{1}} = \frac{l}{l + 2 x_{1}}, k = 2$

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