Waves

JEE Main level, single correct, medium

19 of 142

Question 19 of 142

A string of length 1m1m and mass 5g5g is fixed at both ends. The tension in the string is 8.0N8.0N. The string is set into vibration using an external vibrator of frequency 100Hz100Hz. The separation between successive nodes on the string is close to

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Answer: (A) 20.0cm20.0cm

V=fλV=f\lambda

Tμ=fλ\sqrt{\dfrac{T}{\mu }}=f\lambda

85×10−3=100λ\sqrt{\dfrac{8}{5\times {10}^{-3}}}=100\lambda

λ=40cm\lambda =40 cm

λ2=20cm\dfrac{\lambda }{2}=20cm

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