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Concepts Of Physics MCQ Edition [Volume 1]
Exam
Concepts Of Physics MCQ Edition [Volume 1]
Subject
Physics
Chapter
Wave Motion and Waves on a String

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Wave Motion and Waves on a String

A wave with a frequency of 200 Hz and an amplitude of 1 mm travels...

A wave with a frequency of 200 Hz200 \text{ Hz} and an amplitude of 1 mm1 \text{ mm} travels on a long string. The string has a linear mass density of 6 g m−16 \text{ g m}^{-1} and is subjected to a tension of 60 N60 \text{ N}. Determine the average power transmitted across a given point on the string.

Options

  1. A0.94 W0.94 \text{ W}
  2. B0.47 W0.47 \text{ W}
  3. C4.7 W4.7 \text{ W}
  4. D0.24 W0.24 \text{ W}

Correct answer

B. 0.47 W0.47 \text{ W}

Step-by-step solution

Given:
Frequency f=200 Hzf = 200 \text{ Hz}
Amplitude A=1 mm=10−3 mA = 1 \text{ mm} = 10^{-3} \text{ m}
Linear mass density μ=6 g m−1=6×10−3 kg m−1\mu = 6 \text{ g m}^{-1} = 6 \times 10^{-3} \text{ kg m}^{-1}
Tension T=60 NT = 60 \text{ N}

The wave speed vv is given by:
v=Tμ=606×10−3=10000=100 m s−1v = \sqrt{\dfrac{T}{\mu}} = \sqrt{\dfrac{60}{6 \times 10^{-3}}} = \sqrt{10000} = 100 \text{ m s}^{-1}

The angular frequency ω\omega is:
ω=2πf=2π×200=400π rad s−1\omega = 2\pi f = 2\pi \times 200 = 400\pi \text{ rad s}^{-1}

The average power transmitted across a given point on the string is:
Pavg=12μvω2A2P_{avg} = \dfrac{1}{2} \mu v \omega^2 A^2

Substituting the values:
Pavg=12×(6×10−3)×100×(400π)2×(10−3)2P_{avg} = \dfrac{1}{2} \times (6 \times 10^{-3}) \times 100 \times (400\pi)^2 \times (10^{-3})^2

Pavg=0.3×160000π2×10−6P_{avg} = 0.3 \times 160000\pi^2 \times 10^{-6}

Pavg=0.048π2P_{avg} = 0.048 \pi^2

Using π2≈9.87\pi^2 \approx 9.87:
Pavg≈0.048×9.87≈0.47 WP_{avg} \approx 0.048 \times 9.87 \approx 0.47 \text{ W}
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