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Waves question

2009 · Q171
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Waves question

2009 · Q171

NEETPhysicsWavesMCQ+4 / −1
Each of the two strings of length 51.6 cm and 49.1 cm are tensioned separately by 20 N force. Mass per unit length of both the strings is same and equal to 1 g/m. When both the strings vibrate simultaneously the number of beats is
  1. A
    7
  2. B
    8
  3. C
    3
  4. D
    5
View written solutionFree

Correct answer: A

l1l_1l1​ = 0.516 m, l2l_2l2​ = 0.491 m, T = 20 N.

Mass per unit length, μ\mu μ = 0.001 kg/m.

Frequency, υ=12lTμ\upsilon = {1 \over {2l}}\sqrt {{T \over \mu }} υ=2l1​μT​​

υ1=12×0.516200.001{\upsilon _1} = {1 \over {2 \times 0.516}}\sqrt {{{20} \over {0.001}}} υ1​=2×0.5161​0.00120​​

υ2=12×0.491200.001{\upsilon _2} = {1 \over {2 \times 0.491}}\sqrt {{{20} \over {0.001}}} υ2​=2×0.4911​0.00120​​

∴\therefore∴ Number of beats = υ1−υ2=7{\upsilon _1} - {\upsilon _2} = 7υ1​−υ2​=7.

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