Wave Motion 2 Question 40

40. A 20cm long string, having a mass of 1.0g, is fixed at both the ends. The tension in the string is 0.5N. The string is set into vibration using an external vibrator of frequency 100Hz. Find the separation (in cm ) between the successive nodes on the string.

(2009)

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Answer:

Correct Answer: 40. (5)

Solution:

  1. Distance between the successive nodes,

d=λ2=v2f=T/μ2f

Substituting the values we get, d=5cm



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