An object of mass m is suspended at the end of a massless wire of length L and area of cross-selection, A. Young modulus of the material of the wire is Y. If the mass is pulled down slightly its frequency of oscillation along the vertical direction is:

Option 1 - f = (1/2π)√(mL/YA)

Option 2 - f = (1/2π)√(YA/mL)

Option 3 - f = (1/2π)√(mA/YL)

Option 4 - f = (1/2π)√(YL/mA)

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Answered by

Vishal Baghel | Contributor-Level 10

4 months ago

Correct Option - 2

Detailed Solution:

f = (1/2π)√ (k/m), k=YA/L. f= (1/2π)√ (YA/mL)

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