Question

# A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of 45 Hz. The mass of the wire is 3.5 × 10¯² kg and its linear mass density is 4.0 × 10¯² kg m¯¹. What is (a) the speed of a transverse wave on the string, and (b) the tension in the string?

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Solution

## Given, the frequency of the fundamental mode with which the wire vibrates is 45 Hz, the mass of the wire is 3.5×10−2  kg and the linear mass density is 4.0×10−2 kg/m. a) The formula to calculate the total length of the wire is, l=M/m Here, the mass of the wire is M and the mass per unit length of the wire is m. Substituting the values in the above equation, we get: l=3.5×10−24.0×10−2=0.875 m The wavelength of the stationary wave (λ) is related to the length of the wire by the relation:λ=2l/nwhere,n= Number of nodes in the wireFor fundamental node, n=1:λ=2lλ=2×0.875=1.75 mThe speed of the transverse wave in the string is given as:v=fλ=45×1.75=78.75 m/s. b) The tension produced in the string is given by the relation:T=v2m=(78.75)2×4.0×10−2=248.06 N Thus, the tension in the string is 248 N.

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