First slide

Stationary waves

Question

A wire length L and mass per unit length $6.0 \times 10^{- 3} k g m^{- 1}$ is put under tension of $540 N$ . Two consecutive frequencies that it resonates at are: $420 H z$ and $490 H z$ . Then L in meters is

Moderate

A

$1.1 m$
B

$2.1 m$
C

$5.1 m$
D

$8.1 m$

Solution

The difference in the frequencies of two successive harmonics is equal to fundamental frequency $n = 490 - 420$ $= 70 H z$
$\begin{array}{l} n = \frac{1}{2 L} \sqrt{\frac{T}{μ}} \\ \Rightarrow 70 = \frac{1}{2 \times L} \sqrt{\frac{540}{6 \times 10^{- 3}}} \\ \Rightarrow L = \frac{1}{140} \times 300 = 2.142 m \end{array}$

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