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Waves question

2010 · Shift 0 · Q70
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Waves question

2010 · Shift 0 · Q70

JEE MainPhysicsWavesMCQ+4 / −1
The equation of a wave on a string of linear mass density 0.04  kg m−10.04\,\,kg\,{m^{ - 1}}0.04kgm−1 is given by y=0.02(m) sin⁡[2π(t0.04(s)−x0.50(m))].y = 0.02\left( m \right)\,\sin \left[ {2\pi \left( {{t \over {0.04\left( s \right)}} - {x \over {0.50\left( m \right)}}} \right)} \right].y=0.02(m)sin[2π(0.04(s)t​−0.50(m)x​)]. The tension in the string is
  1. A
    4.0N4.0N4.0N
  2. B
    12.5N12.5N12.5N
  3. C
    0.5N0.5N0.5N
  4. D
    6.25N6.25N6.25N
View written solutionFree

Correct answer: D

  1. Write the given wave equation in standard form

The wave is

y=0.02sin⁡[2π(t0.04−x0.50)].y = 0.02\sin\left[2\pi\left(\frac{t}{0.04}-\frac{x}{0.50}\right)\right].y=0.02sin[2π(0.04t​−0.50x​)].

A progressive wave can be written as

y=Asin⁡[2π(tT−xλ)].y = A\sin\left[2\pi\left(\frac{t}{T}-\frac{x}{\lambda}\right)\right].y=Asin[2π(Tt​−λx​)].

Comparing, we get:

  • Amplitude: A=0.02 mA = 0.02\,\text{m}A=0.02m
  • Time period: T=0.04 sT = 0.04\,\text{s}T=0.04s
  • Wavelength: λ=0.50 m\lambda = 0.50\,\text{m}λ=0.50m
  1. Find the wave speed

Wave speed is

v=λT=0.500.04=12.5 m s−1.v = \frac{\lambda}{T} = \frac{0.50}{0.04} = 12.5\,\text{m s}^{-1}.v=Tλ​=0.040.50​=12.5m s−1.
  1. Use the relation between speed and tension

For a wave on a stretched string,

v=Tμ,v = \sqrt{\frac{T}{\mu}},v=μT​​,

where:

  • TTT = tension in the string
  • μ=0.04 kg m−1\mu = 0.04\,\text{kg m}^{-1}μ=0.04kg m−1

So,

T=μv2.T = \mu v^2.T=μv2.

Substitute the values:

T=0.04×(12.5)2T = 0.04 \times (12.5)^2T=0.04×(12.5)2 =0.04×156.25= 0.04 \times 156.25=0.04×156.25 =6.25 N.= 6.25\,\text{N}.=6.25N.
  1. Check options
  • A: 4.0 N4.0\,\text{N}4.0N
  • B: 12.5 N12.5\,\text{N}12.5N
  • C: 0.5 N0.5\,\text{N}0.5N
  • D: 6.25 N6.25\,\text{N}6.25N

Hence the correct option is D.

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