One end of a long metallic wire of length L is tied to the ceiling. The other end is tied to a massless spring of spring constant k . A mass m hangs freely from the free end of the spring. The area of cross-section and Young’s modulus of the wire are A and Y respectively. If the mass is slightly pulled down and released, it will oscillate with a time period T equal to

Question

One end of a long metallic wire of length L is tied to the ceiling. The other end is tied to a massless spring of spring constant k . A mass m hangs freely from the free end of the spring. The area of cross-section and Young’s modulus of the wire are A and Y respectively. If the mass is slightly pulled down and released, it will oscillate with a time period T equal to, 2 π m k 1 / 2, 2 π m Y A + k L Y A k, 2 π m Y A k L 1 2, 2 π m L Y A 1 2

MARKS App · Accepted Answer

Equivalent force constant for a wire is given by k = Y A L . Because in case of a wire, F = Y A L ∆ L and in case of spring, F = k ∆ x . Comparing these two, we find k of wire = Y A L k eq = k 1 k 2 k 1 + k 2 = Y A L k Y A L + k = Y A k Y A + L k ∴ T = 2 π m k eq = 2 π m Y A + L k Y A k

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