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Q.

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

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a

15.0 cm

b

33.3 cm

c

16.6 cm

d

20.0 cm

answer is D.

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Detailed Solution

Velocity of wave, v=Tμ

=85×10−31=8×1035=40m/s

Wavelength of wave v = fλ

λ=vf=40100m

Therefore, distance between two

consecutive nodes =λ2

=40100×2=20100=20cm

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