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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

One end of a string having a total length of 30 cm and a linear...

One end of a string having a total length of 30 cm30 \text{ cm} and a linear mass density of 0.5 g cm−10.5 \text{ g cm}^{-1} is fixed to a wall, and the other end is attached to a frictionless ring that slides on a vertical rod. A wave pulse moves along the string towards the ring at a speed of 20 cm s−120 \text{ cm s}^{-1}. The pulse is symmetric about its maximum, initially located 20 cm20 \text{ cm} from the ring. Assuming no energy is lost during reflections at the ends, what is the tension in the string?

Options

  1. A5×10−4 N5 \times 10^{-4} \text{ N}
  2. B4×10−3 N4 \times 10^{-3} \text{ N}
  3. C2×10−3 N2 \times 10^{-3} \text{ N}
  4. D2×10−2 N2 \times 10^{-2} \text{ N}

Correct answer

C. 2×10−3 N2 \times 10^{-3} \text{ N}

Step-by-step solution

Given, linear mass density of the string, μ=0.5 g cm−1\mu = 0.5 \text{ g cm}^{-1}.
Converting to SI units:
μ=0.5×10−3 kg10−2 m=0.05 kg m−1\mu = \dfrac{0.5 \times 10^{-3} \text{ kg}}{10^{-2} \text{ m}} = 0.05 \text{ kg m}^{-1}

Speed of the wave pulse, v=20 cm s−1=0.2 m s−1v = 20 \text{ cm s}^{-1} = 0.2 \text{ m s}^{-1}.

The speed of a transverse wave on a stretched string is given by the relation:
v=Tμv = \sqrt{\dfrac{T}{\mu}}

Rearranging the formula to solve for tension TT:
T=μv2T = \mu v^2

Substituting the given values:
T=0.05×(0.2)2T = 0.05 \times (0.2)^2
T=0.05×0.04T = 0.05 \times 0.04
T=0.002 N=2×10−3 NT = 0.002 \text{ N} = 2 \times 10^{-3} \text{ N}
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