Stationary waves

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A wire of length $l$ and mass per unit length $6.0 \times 10^{- 3} k g / m$ 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 _____

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Solution

$\frac{n v}{2 l} = 420 \to (i) \frac{(n + 1) v}{2 l} = 490 \to (i i) D i v i d e i / i i n = 6 S u b s t i t u t e i n e q u a t i o n (i) w e h a v e \Rightarrow \frac{6 \times v}{2 l} = 420 A l s o, v = \sqrt{\frac{T}{μ}} = \sqrt{\frac{540}{6 \times 10^{- 3}}} ∴ \frac{6 \times \sqrt{\frac{540}{6 \times 10^{- 3}}}}{2 l} = 420 l = \frac{300}{140} = 2.142$

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Stationary waves

Remember concepts with our Masterclasses.

A wire of length l and mass per unit length 6.0×10−3 kg/m is put under tension of 540N two consecutive frequencies that it resonates at are 420 Hz and 490 Hz , then l in meters is _____

nv2l=420 → i n+1v2l=490 → (ii) Divide i/ii n=6 Substitute in equation (i) we have ⇒6×v2l=420 Also, v=Tμ=5406×10−3 ∴6×5406×10−32l=420 l=300140=2.142

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A wire of length $l$ and mass per unit length $6.0 \times 10^{- 3} k g / m$ 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 _____