Question

A stretched sonometer wire is in unison with a tuning fork when it is stretched by certain weight in air corresponding resonating length is 100 cm if the weight is completely immersed in water the resonating length of a wire decreases to 90 cm find the density of material of weight

Answer 1

Answer:

Density of the material is 5263.16 kg/m^3

Explanation:

As we know that the frequency of standing wave in wire of length L is given as

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

so here when weight is suspended in air

$f = \frac{1}{2(100)}\sqrt{\frac{Mg}{\mu}}$

now when it is immersed in water

$f = \frac{1}{2(90)}\sqrt{\frac{Mg(1 - \frac{\rho_w}{\rho_o})}{\mu}}$

now from above two equations we have

$\frac{1}{2(100)}\sqrt{\frac{Mg}{\mu}} = \frac{1}{2(90)}\sqrt{\frac{Mg(1 - \frac{\rho_w}{\rho_o})}{\mu}}$

$\frac{1}{100} = \frac{1}{90}\sqrt{(1 - \frac{\rho_w}{\rho_o})}$

$0.81 = 1 - \frac{\rho_w}{\rho_o}$

$\frac{\rho_w}{\rho_o} = 0.19$

$\rho_o = 5263.16 kg/m^3$

#Learn

Topic : Standing waves

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