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A 1m long wire having tension of 100 N and of linear mass density 0.01 kg/m is fixed and A and free at end B. the point C which is 20cm from end B is constrained to be stationary. To create resonance in the wire, the minimum frequency of the tuning fork will be : A. 125 HzB. 150 HzC. 300 HzD. 275 Hz |
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Answer» Correct Answer - A Length of BC to AC is `(20)/(80)=(1)/(4)` So, value of loops in BC to AC will also be `1 : 4` `f_(1)=(1)/(4(0.2))sqrt((T)/(mu))=125 Hz` `f_(2) = (3)/(4(0.2))sqrt((T)/(mu))=275 Hz`. |
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