Physics - Wave on String Question with Solution | TestHub

PhysicsWave on StringMiscellaneousMedium2 minPYQ_2021
PhysicsMediumnumerical

A wire having a linear mass density9.0×10-4 kg m-1is stretched between two rigid supports with a tension of900 N.The wire resonates at a frequency of500 Hz.The next higher frequency at which the same wire resonates is550 Hz.The length of the wire is ___________m.

Answer:
10.00
Solution:

fn=nv2=500
fn+1=(n+1)v2=550
n+1n=1110
Thus,n=10
Thus,=nv2fnv=Tμ
=102×500×9009×10-4=10 m

Stream:JEESubject:PhysicsTopic:Wave on StringSubtopic:Miscellaneous
2mℹ️ Source: PYQ_2021

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