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Waves question

2019 · 11 Jan · Shift 1 · Q43
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Waves question

2019 · 11 Jan · Shift 1 · Q43

JEE MainPhysicsWavesMCQ+4 / −1
Equation of travelling wave on a stretched string of linear density 5 g/m is y = 0.03 sin(450 t – 9x) where distance and time are measured in SI units. The tension in the string is :
  1. A
    10 N
  2. B
    7.5 N
  3. C
    5 N
  4. D
    12.5 N
View written solutionFree

Correct answer: D

  1. Write the standard form of a travelling wave

A wave on a string is generally written as y=Asin⁡(ωt−kx)y = A\sin(\omega t - kx)y=Asin(ωt−kx) where:

  • AAA = amplitude
  • ω\omegaω = angular frequency
  • kkk = wave number

Given: y=0.03sin⁡(450t−9x)y = 0.03\sin(450t - 9x)y=0.03sin(450t−9x) So, ω=450 rad/s,k=9 rad/m\omega = 450\ \text{rad/s}, \qquad k = 9\ \text{rad/m}ω=450 rad/s,k=9 rad/m

  1. Find the wave speed

Wave speed is v=ωk=4509=50 m/sv = \frac{\omega}{k} = \frac{450}{9} = 50\ \text{m/s}v=kω​=9450​=50 m/s

  1. Use the relation between speed and tension

For a stretched string, v=Tμv = \sqrt{\frac{T}{\mu}}v=μT​​ where:

  • TTT = tension in string
  • μ\muμ = linear mass density

Given linear density: μ=5 g/m=5×10−3 kg/m\mu = 5\ \text{g/m} = 5 \times 10^{-3}\ \text{kg/m}μ=5 g/m=5×10−3 kg/m

So, T=μv2T = \mu v^2T=μv2 Substitute values: T=5×10−3×(50)2T = 5 \times 10^{-3} \times (50)^2T=5×10−3×(50)2 T=5×10−3×2500T = 5 \times 10^{-3} \times 2500T=5×10−3×2500 T=12.5 NT = 12.5\ \text{N}T=12.5 N

  1. Match with the options

T=12.5 NT = 12.5\ \text{N}T=12.5 N So the correct option is D.

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