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?

Answer 1

Given, the frequency of the fundamental mode with which the wire vibrates is 45 Hz, the mass of the wire is $3.5\times {10}^{-2}$  kg and the linear mass density is $4.0\times {10}^{-2}$ 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:

$l=\frac{3.5\times {10}^{-2}}{4.0\times {10}^{-2}}=0.875m$

The wavelength of the stationary wave (λ) is related to the length of the wire by the relation:

$\lambda =2l/n$

where,

n= Number of nodes in the wire

For fundamental node, n=1:

$\lambda =2l$

$\lambda =2\times 0.875=1.75m$

The speed of the transverse wave in the string is given as:

$v=f\lambda =45\times 1.75=78.75m/s$.

b)

The tension produced in the string is given by the relation:

$T=v^{2}m=(78.75{)}^2\times 4.0\times {10}^{-2}=248.06N$

Thus, the tension in the string is 248 N.