For a wave represented by the equation y = A cos ( ω t − kx ) , find the energy that crosses the location x = 2 k π during the time interval from t = 6 ω 5 π to t = 6 ω 7 π

Power crossing a point: P = μ ω 2 A 2 v sin 2 ( k x − ω t ) , with v = ω / k . At x = π /2 k , P = k μ ω 3 A 2 cos 2 ( ω t ) . E = k μ ω 3 A 2 ∫ 5 π /6 ω 7 π /6 ω cos 2 ( ω t ) d t = k μ ω 2 A 2 [ 2 s + 4 s i n 2 s ] 5 π /6 7 π /6 . = k μ ω 2 A 2 ( 6 π + 4 3 ) = 12 k μ ω 2 A 2 ( 3 3 + 2 π ) - (C).

Physics - Wave on String Question with Solution | TestHub

PhysicsWave on StringPower EnergyHard2 minQB

PhysicsHardsingle choice

Passage / Comprehension

In a transverse wave on a string, the kinetic energy is due to the oscillation of the particles, and the potential energy is due to the tension stretching the string. The power is the energy transmitted per unit time. The inclination at every point on the string with respect to its initial direction is small. The linear density of the string is $μ$ .

For a wave represented by the equation $y = A cos (ω t - kx)$ , find the energy that crosses the location $x = \frac{π}{2 k}$ during the time interval from $t = \frac{5 π}{6 ω}$ to $t = \frac{7 π}{6 ω}$

Options:

Answer:

Solution:

Power crossing a point: $P = μ ω^{2} A^{2} v sin^{2} (k x - ω t)$ , with $v = ω / k$ .

At $x = π /2 k$ , $P = \frac{μ ω ^{3} A ^{2}}{k} cos^{2} (ω t)$ .

$E = \frac{μ ω ^{3} A ^{2}}{k} \int_{5 π /6 ω}^{7 π /6 ω} cos^{2} (ω t) d t = \frac{μ ω ^{2} A ^{2}}{k} [\frac{s}{2} + \frac{s i n 2 s}{4}]_{5 π /6}^{7 π /6}$ .

$= \frac{μ ω ^{2} A ^{2}}{k} (\frac{π}{6} + \frac{3}{4}) = \frac{μ ω ^{2} A ^{2}}{12 k} (33 + 2 π)$ - (C).

Stream:JEESubject:PhysicsTopic:Wave on StringSubtopic:Power Energy

⏱ 2mℹ️ Source: QB

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