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Waves question

2019 · 9 Apr · Shift 1 · Q53
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Waves question

2019 · 9 Apr · Shift 1 · Q53

JEE MainPhysicsWavesMCQ+4 / −1
A string is clamped at both the ends and it is vibrating in its 4th harmonic. The equation of the stationary wave is Y = 0.3 sin(0.157x) cos(200pt). The length of the string is : (All quantities are in SI units.)
  1. A
    60 m
  2. B
    20 m
  3. C
    80 m
  4. D
    40 m
View written solutionFree

Correct answer: C

  1. Compare with the standard form of a stationary wave

    The given equation is Y=0.3sin⁡(0.157x)cos⁡(200πt).Y = 0.3\sin(0.157x)\cos(200\pi t).Y=0.3sin(0.157x)cos(200πt).

    The standard form of a stationary wave on a string is y=Asin⁡(kx)cos⁡(ωt),y = A\sin(kx)\cos(\omega t),y=Asin(kx)cos(ωt), where kkk is the wave number.

    Hence, k=0.157 rad m−1.k = 0.157\ \text{rad m}^{-1}.k=0.157 rad m−1.

  2. Relate wave number to wavelength

    We know, k=2πλ.k = \frac{2\pi}{\lambda}.k=λ2π​.

    Therefore, λ=2πk=2π0.157.\lambda = \frac{2\pi}{k} = \frac{2\pi}{0.157}.λ=k2π​=0.1572π​.

    Using 2π≈6.2832\pi \approx 6.2832π≈6.283, λ≈6.2830.157≈40 m.\lambda \approx \frac{6.283}{0.157} \approx 40\ \text{m}. λ≈0.1576.283​≈40 m.

  3. Use the condition for harmonics in a string fixed at both ends

    For a string fixed at both ends, in the nnn-th harmonic, L=nλ2.L = \frac{n\lambda}{2}.L=2nλ​.

    Since the string is vibrating in its 4th harmonic, n=4n=4n=4.

    So, L=4λ2=2λ.L = \frac{4\lambda}{2} = 2\lambda.L=24λ​=2λ.

    Substituting λ=40 m\lambda = 40\ \text{m}λ=40 m, L=2×40=80 m.L = 2 \times 40 = 80\ \text{m}. L=2×40=80 m.

  4. Match with the options

    L=80 m.L = 80\ \text{m}.L=80 m.

    So the correct option is C.

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