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  • Help me understand! A string of length 1 m and mass 5 g 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 100 Hz. The separation between successive nodes on the string is clo

A string of length 1 m and mass 5 g 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 100 Hz. The separation between successive nodes on the string is close to: 

  • Option 1)

  • Option 2)

  • Option 3)

  • Option 4)

Answers (1)

best_answer

 

Speed of wave on string -

v= \sqrt{\frac{T}{\mu }}
 

- wherein

T= Tension in the string

\mu = linear mass density

Given

m=5g=5\times 10^{-3}Kg\\\\L=1m

so\: \mu =\frac{M}{L}=5\times 10^{-3}\\\\velocity \: of\: wave\: on\: \: string\\V=\sqrt{\frac{T}{\mu }}=\sqrt{\frac{8}{5\times 10^{-3}}} = 40m/s\\\\now \: for \: wave \\\\\lambda =\frac{V}{f}\\\\\\f=100Hz\\\\so\: \: \lambda =\frac{40}{100}=0.4m\\\\d=\frac{\lambda }{2}\\\\d=\frac{40}{2\times 100}=20cm

 

 


Option 1)

Option 2)

Option 3)

Option 4)

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