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A 200 Hz wave with an amplitude 1 mm travels on a long string of linear mass density $6 g m^{- 1}$ kept under a tension of 60 N. the average power transmitted across a given point on the string is $P_{a v}$ and the total energy associated with the wave in a 2m long portion of the string is E. Then,

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$P_{a v} = 0.47 W$

$E = 9.4 m J$

$P_{a v} = 0.94 W$

$E = 18.8 m J$

answer is A, B.

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Detailed Solution

Given,

$f = 200 H z, A = 10^{- 3} m, μ$

= $0.006 k g m^{- 1}, T = 60 N$

$P_{a v} = \frac{1}{2} μ ω^{2} A^{2} v$

$= \frac{1}{2} (0.006) {(2 π \times 200)}^{2} {(10^{- 3})}^{2} \sqrt{\frac{60}{0.006}}$

Or $P_{a v} = 0.47 W$

Now,

$E = \frac{1}{2} m ω^{2} A^{2}$

$= \frac{1}{2} (0.006 \times 2) {(2 π \times 200)}^{2} {(10^{- 3})}^{2}$

Or $E = 9.4 m J$ .

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