Question

Question: Hot chocolate effectTap a metal spoon inside a mug of water and note the frequency f_iyou hear. Then add a spoonful of powder (say, chocolate mix or instant coffee) and tap again as you stir the powder. The frequency you hear has a lower value f_sbecause the tiny air bubbles released by the powder change the water’s bulk modulus. As the bubbles reach the water surface and disappear, the frequency gradually shifts back to its initial value. During the effect, the bubbles don’t appreciably change the water’s density or volume of dV/dp- that is , the differential change in volume due to the differential change in the pressure caused by the sound wave in the water . If${\mathbf{f}}_{\mathbf{s}}\mathbf{/}{\mathbf{f}}_{\mathbf{i}}$=0.333, what is the ratio${\left(\frac{\mathbf{dV}}{\mathbf{dp}}\right)}_{\mathbf{s}}\mathbf{/}{\left(\frac{\mathbf{dV}}{\mathbf{dp}}\right)}_{\mathbf{i}}$?

Short answer

Answer

The ratio is, $\frac{{\left(dp/dV\right)}_s}{{\left(dp/dV\right)}_i}=9.00$.

Step-by-step solution

Step 1: Given data

The ratio is, $\frac{f_s}{f_i}=0.333$.

Determining the concept

The expression for the speed of the sound in terms of frequency and wavelength is given by,

$v=\lambda f$…… (i)

Here v is the speed of the sound, $\lambda$is the wavelength and f is the frequency

The speed of the sound in terms of bulk modulus and density is given by,

$v=\sqrt{\frac{B}{\rho }}$…… (ii)

Here B is the bulk modulus and $\rho$is the density.

The expression for bulk modulus is given by,

localid="1662381277405" $B=-\frac{dp/dV}{V}$…… (iii)

Determining the ratio

Since, v,$\lambda and\rho$ are not changed appreciably, the frequency ratio becomes,

$$\begin{gathered} \frac{f_s}{f_i}=\frac{v_s/\lambda }{v_i/\lambda } \\ \Rightarrow \frac{f_s}{f_i}=\frac{v_s}{v_i} \end{gathered}$$

Using the concept of equation (ii) we can write,

$$\begin{gathered} \frac{f_s}{f_i}=\sqrt{\frac{B_s/\rho }{B_i/\rho }} \\ \Rightarrow \frac{f_s}{f_i}=\sqrt{\frac{B_s}{B_i}} \end{gathered}$$

Using the concept of equation (iii) we can write,

$$\begin{gathered} \frac{f_s}{f_i}=\sqrt{\frac{{\left(-\frac{dp/dV}{V}\right)}_s}{{\left(-\frac{dp/dV}{V}\right)}_i}} \\ \Rightarrow \frac{f_s}{f_i}=\sqrt{\frac{{\left(dp/dV\right)}_s}{{\left(dp/dV\right)}_i}} \end{gathered}$$

But,
$\frac{f_s}{f_i}=\sqrt{\frac{B_s}{B_i}}$

Thus,

$$\begin{gathered} \frac{f_s}{f_i}=\sqrt{\frac{B_s}{B_i}} \\ =\sqrt{\frac{{\left(dp/dV\right)}_s}{{\left(dp/dV\right)}_i}} \end{gathered}$$

Squaring,

$$\begin{gathered} {\left(\frac{f_s}{f_i}\right)}^2=\frac{B_s}{B_i} \\ =\frac{{\left(dp/dV\right)}_s}{{\left(dp/dV\right)}_i} \end{gathered}$$

Thus,

$\frac{{\left(dp/dV\right)}_s}{{\left(dp/dV\right)}_i}={\left(\frac{f_s}{f_i}\right)}^2$

It is given that the,$\frac{f_s}{f_i}=0.333$,

$\therefore \frac{{\left(dp/dV\right)}_s}{{\left(dp/dV\right)}_i}=9.00$

Hence, the ratio is,$\frac{{\left(dp/dV\right)}_s}{{\left(dp/dV\right)}_i}=9.00$ .