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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 transverse wave given by y = (0.02 m) [(1.0 m^ -1 )x + (30 s^ -1...

A transverse wave given by y=(0.02 m)sin⁡[(1.0 m−1)x+(30 s−1)t]y = (0.02 \text{ m}) \sin[(1.0 \text{ m}^{-1})x + (30 \text{ s}^{-1})t] propagates along a stretched string with a linear mass density of 1.2×10−4 kg m−11.2 \times 10^{-4} \text{ kg m}^{-1}. Determine the tension in the string.

Options

  1. A0.108 N0.108 \text{ N}
  2. B0.012 N0.012 \text{ N}
  3. C0.0036 N0.0036 \text{ N}
  4. D1.08 N1.08 \text{ N}

Correct answer

A. 0.108 N0.108 \text{ N}

Step-by-step solution

Comparing the given equation y=0.02sin⁡(x+30t)y = 0.02 \sin(x + 30t) with the standard wave equation y=Asin⁡(kx+ωt)y = A \sin(kx + \omega t), we get:

k=1.0 m−1k = 1.0 \text{ m}^{-1}

ω=30 s−1\omega = 30 \text{ s}^{-1}

The wave speed vv is given by:

v=ωk=301.0=30 m s−1v = \dfrac{\omega}{k} = \dfrac{30}{1.0} = 30 \text{ m s}^{-1}

The wave speed on a stretched string is related to tension TT and linear mass density μ\mu by:

v=Tμ⇒T=μv2v = \sqrt{\dfrac{T}{\mu}} \Rightarrow T = \mu v^2

Given μ=1.2×10−4 kg m−1\mu = 1.2 \times 10^{-4} \text{ kg m}^{-1}, substituting the values:

T=(1.2×10−4)×(30)2T = (1.2 \times 10^{-4}) \times (30)^2

T=1.2×10−4×900T = 1.2 \times 10^{-4} \times 900

T=1080×10−4=0.108 NT = 1080 \times 10^{-4} = 0.108 \text{ N}
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