A sign is hanging from a single metal wire. The shop owner notices that the wire

Question

A sign is hanging from a single metal wire. The shop owner notices that the wire vibrates at a fundamental resonance frequency of 679 Hz, which irritates his customers. In an attempt to fix the problem, the shop owner cuts the wire in half and hangs the sign from the two halves. Assuming the tension in the two wires to be the same, what is the new fundamental frequency of each wire?

Explanation / Answer

let the mass of the signboard be m. let length of the string be l. then in first case T=m*g in second case, 2*T=m*g ==>T=m*g/2 so as frequency is proportional to square root of tension in the wire, f1/f2=sqrt(mg)/sqrt(m*g/2) ==>679/f2=sqrt(2) f2=679/sqrt(2)=480.125 Hz

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