Question

A 1.80-m -long uniform bar that weighs 531 N is suspended in a horizontal position by two vertical wires that are attached to the ceiling. One wire is aluminum and the other is copper. The aluminum wire is attached to the left-hand end of the bar, and the copper wire is attached 0.40 m to the left of the right-hand end. Each wire has length 0.600 m and a circular cross section with radius 0.250 mm .a. What is the fundamental frequency of transverse standing waves for aluminium wire? b. What is the fundamental frequency of transverse standing waves for copper wire?

Answer 1

Answer:

(a) 498.4 Hz

(b) 442 Hz

Solution:

As per the question:

Length of the wire, L = 1.80 m

Weight of the bar, W = 531 N

The position of the copper wire from the left to the right hand end, x = 0.40 m

Length of each wire, l = 0.600 m

Radius of the circular cross-section, R = 0.250 mm = $.250\times 10^{- 3}\ m$

Now,

Applying the equilibrium condition at the left end for torque:

$T_{Al}.0 + T_{C}(L - x) = W\frac{L}{2}$

$T_{C}(1.80 - 0.40) = 531\times \frac{1.80}{2}$

$T_{C} = 341.357\ Nm$

The weight of the wire balances the tension in both the wires collectively:

$W = T_{Al} + T_{C}$

$531 = T_{Al} + 341.357$

$T_{Al} = 189.643\ Nm$

Now,

The fundamental frequency is given by:

$f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$

where

$\mu = A\rho = \pi R^{2}\rho$

(a) For the fundamental frequency of Aluminium:

$f = \frac{1}{2L}\sqrt{\frac{T_{Al}}{\mu}}$

$f = \frac{1}{2L}\sqrt{\frac{T_{Al}}{\pi R^{2}\rho_{Al}}}$

where

$\rho_{l} = 2.70\times 10^{3}\ kg/m^{3}$

$f = \frac{1}{2\times 0.600}\sqrt{\frac{189.643}{\pi 0.250\times 10^{- 3}^{2}\times 2.70\times 10^{3}}} = 498.4\ Hz$

(b)  For the fundamental frequency of Copper:

$f = \frac{1}{2L}\sqrt{\frac{T_{C}}{\mu}}$

$f = \frac{1}{2L}\sqrt{\frac{T_{C}}{\pi R^{2}\rho_{C}}}$

where

$\rho_{C} = 8.90\times 10^{3}\ kg/m^{3}$

$f = \frac{1}{2\times 0.600}\sqrt{\frac{341.357}{\pi 0.250\times 10^{- 3}^{2}\times 2.70\times 10^{3}}} = 442\ Hz$