Question
Download Solution PDFA rod of mass 'M' and length '2L' is suspended at its middle by a wire. It exhibits torsional oscillations; if two masses each of 'm' are attached at distance 'L/2' from its centre of both sides, it reduces the oscillation frequency by 10%. The value of the ratio M/m is close to :
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
The frequency of a Torsional Pendulum is given by
where K = Torsional Constant, I = Moment of Inertia
Calculation:
Given:
Mass of the rod = M, Length of the rod = 2L, Mass of 2 attached masses = m, Location of attachment = L/2 from center of rod of both sides,
Reduction in frequency = 10 %
⇒
⇒ f2 = 0.9f1 ................. (i)
Case 1:
Moment of Inertia,
Frequency of Oscillation,
Case 2:
Moment of Inertia,
Frequency of Oscillation,
From equation (i), we have
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