A string of length 1m and mass 5g is fixed at both ends. The tension in the string is 8 N . The string is sent into vibration using an external vibrator of frequency 100 Hz . The separation between successive nodes on the string is close to

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16.6 cm

20.0 cm

10.0 cm

33.3 cm

answer is B.

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Detailed Solution

Velocity of wave on string V=Tμ=85×10−3=40 m/sec Wave length of wave : λ=vf ⇒λ=40100m Separation between successive nodes =λ/2 = 20100=0.2 m =20 cm

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