First slide

Stationary waves

Question

A wire having a linear density of $0.05 g / c m$ is stretched between two rigid supports with a tension of $4.5 \times 10^{2} N$ it is observed that the wire resonance at a frequency of 420 Hz. The next higher frequency at which the wire resonates is 490 Hz. Determine the length of the wire.

Moderate

A

2.14 m
B

3.2 m
C

4 m
D

8 m

Solution

$E x p r e s s i o n f o r F r e q u e n c y w h e n b o t h e n d s a r e t i e d f = \frac{n}{2 L} \sqrt{\frac{T}{μ}} N o w, 420 = \frac{n}{2 L} \sqrt{\frac{T}{μ}} \to (i) 490 = \frac{n + 1}{2 L} \sqrt{\frac{T}{μ}} \to (i i) \begin{array}{l} D i v i d e e q u a t i o n i \div i i \\ n = 6 \\ P u t t i n g i n e q u a t i o n (i) \\ \begin{array}{l} 420 = \frac{6}{2 L} \sqrt{\frac{4.5 \times 10^{2}}{0.05 \times \frac{10^{- 3}}{10^{- 2}}}} \\ \Rightarrow L = 2.14 m \end{array} \end{array}$

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