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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 4.0 kg block is suspended from the ceiling of an elevator by a...

A 4.0 kg4.0 \text{ kg} block is suspended from the ceiling of an elevator by a string with a linear mass density of 19.2×10−3 kg m−119.2 \times 10^{-3} \text{ kg m}^{-1}. Determine the speed (relative to the string) at which a wave pulse can travel along the string, given that the elevator accelerates upward at a rate of 2.0 m s−22.0 \text{ m s}^{-2}. Use g=10 m s−2g = 10 \text{ m s}^{-2}.

Options

  1. A41 m s−141 \text{ m s}^{-1}
  2. B20 m s−120 \text{ m s}^{-1}
  3. C50 m s−150 \text{ m s}^{-1}
  4. D46 m s−146 \text{ m s}^{-1}

Correct answer

C. 50 m s−150 \text{ m s}^{-1}

Step-by-step solution

The effective acceleration of the block in the elevator accelerating upward is given by:

geff=g+a=10+2.0=12.0 m s−2g_{\text{eff}} = g + a = 10 + 2.0 = 12.0 \text{ m s}^{-2}

The tension in the string is equal to the apparent weight of the suspended block:

T=Mgeff=4.0×12.0=48 NT = M g_{\text{eff}} = 4.0 \times 12.0 = 48 \text{ N}

The speed of a transverse wave pulse on a string is given by the formula:

v=Tμv = \sqrt{\dfrac{T}{\mu}}

Substituting the given values:

v=4819.2×10−3v = \sqrt{\dfrac{48}{19.2 \times 10^{-3}}}

v=4800019.2=2500v = \sqrt{\dfrac{48000}{19.2}} = \sqrt{2500}

v=50 m s−1v = 50 \text{ m s}^{-1}
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