An object of specific gravity ρ is hung from a thin steel wire. The fundamental frequency for transverse standing waves in the wire is 300 Hz . The object is immersed in water so that one half of its

Question

An object of specific gravity ρ is hung from a thin steel wire. The fundamental frequency for transverse standing waves in the wire is 300 Hz . The object is immersed in water so that one half of its volume is submerged. The new fundamental frequency (in Hz ) is:

TestHub · Accepted Answer

The diagrammatic representation of the given problem is shown in the figure. The expression of the fundamental frequency is $v = \frac{1}{2 l} \sqrt{\frac{T}{μ}}$

$T = m g = v ρ^{'} g$ ; $f = \frac{1}{2 l} \sqrt{\frac{v ρ^{'} g}{μ}}$

$T^{'} = m g - \frac{V}{2} ρ_{w} g$

$f^{'} = \frac{1}{2 l} \sqrt{\frac{v ρ^{'} - \frac{v}{2} ρ_{w} g}{μ}}$

$\frac{f^{'}}{f} = \sqrt{\frac{2 ρ^{'} - ρ_{w}}{2 ρ^{'}}} \frac{ρ^{'}}{ρ_{w}} = s p e c i f i c g r a v i t y = ρ$

$f^{'} = 300 \sqrt{\frac{2 ρ - 1}{2 ρ}} H z$

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