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Waves question

2019 · 10 Jan · Shift 1 · Q49
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Waves question

2019 · 10 Jan · Shift 1 · Q49

JEE MainPhysicsWavesMCQ+4 / −1
A string of length 1 m and mass 5 g is fixed at both ends. The tension in the string is 8.0 N. The string is set into vibration using an external vibrator of frequency 100 Hz. The separation between successive nodes on the string is close to -
  1. A
    16.6 cm
  2. B
    10.0 cm
  3. C
    20.0 cm
  4. D
    33.3 cm
View written solutionFree

Correct answer: C

  1. Given data
  • Length of string: L=1 mL = 1\,\text{m}L=1m
  • Mass of string: m=5 g=0.005 kgm = 5\,\text{g} = 0.005\,\text{kg}m=5g=0.005kg
  • Tension: T=8.0 NT = 8.0\,\text{N}T=8.0N
  • Driving frequency: f=100 Hzf = 100\,\text{Hz}f=100Hz
  1. Find linear mass density

μ=mL=0.0051=0.005 kg m−1\mu = \frac{m}{L} = \frac{0.005}{1} = 0.005\,\text{kg m}^{-1}μ=Lm​=10.005​=0.005kg m−1

  1. Find wave speed on the string

For a stretched string,

v=Tμv = \sqrt{\frac{T}{\mu}}v=μT​​

So,

v=8.00.005=1600=40 m/sv = \sqrt{\frac{8.0}{0.005}} = \sqrt{1600} = 40\,\text{m/s}v=0.0058.0​​=1600​=40m/s

  1. Find wavelength corresponding to frequency 100 Hz100\,\text{Hz}100Hz

Using

v=fλv = f\lambdav=fλ

λ=vf=40100=0.4 m\lambda = \frac{v}{f} = \frac{40}{100} = 0.4\,\text{m}λ=fv​=10040​=0.4m

  1. Relation between node separation and wavelength

In a standing wave, the distance between two successive nodes is

λ2\frac{\lambda}{2}2λ​

Hence,

node separation=0.42=0.2 m=20.0 cm\text{node separation} = \frac{0.4}{2} = 0.2\,\text{m} = 20.0\,\text{cm}node separation=20.4​=0.2m=20.0cm

  1. Check with options
  • A: 16.6 cm16.6\,\text{cm}16.6cm
  • B: 10.0 cm10.0\,\text{cm}10.0cm
  • C: 20.0 cm20.0\,\text{cm}20.0cm
  • D: 33.3 cm33.3\,\text{cm}33.3cm

Therefore, the correct option is C.

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