Question

A plane progressive transverse wave travels in a medium M₁ and enters into another medium M₂ in which its speed decreases to 60%. Then the ratio of the intensity of the transmitted and the incident waves is n then 64n is:

• A. 36
• B. 48
• C. 60
• D. 72

Answer 1

Answer: (C)

60

Answer 2

Let $v_1$ be the speed of the wave in medium M₁ and $v_2$ be the speed of the wave in medium M₂. According to the problem, the speed in medium M₂ decreases to 60% of the speed in medium M₁. So, $v_2 = 0.60 v_1$.

When a wave travels from one medium to another, the frequency ($\omega$) of the wave remains unchanged. Let $A_i$ be the amplitude of the incident wave and $A_t$ be the amplitude of the transmitted wave. Let $\rho_1$ and $\rho_2$ be the densities of medium M₁ and medium M₂ respectively.

The intensity of a wave is given by $I = \frac{1}{2} \rho \omega^2 A^2 v$. The intensity of the incident wave is $I_i = \frac{1}{2} \rho_1 \omega^2 A_i^2 v_1$. The intensity of the transmitted wave is $I_t = \frac{1}{2} \rho_2 \omega^2 A_t^2 v_2$.

The ratio of the intensity of the transmitted wave and the incident wave is $n$: $n = \frac{I_t}{I_i} = \frac{\frac{1}{2} \rho_2 \omega^2 A_t^2 v_2}{\frac{1}{2} \rho_1 \omega^2 A_i^2 v_1} = \frac{\rho_2 v_2}{\rho_1 v_1} \left(\frac{A_t}{A_i}\right)^2$.

When a wave passes from one medium to another, the amplitude of the transmitted wave ($A_t$) and the incident wave ($A_i$) are related by the transmission coefficient. For mechanical waves, this relationship depends on the acoustic impedances of the media. The acoustic impedance $Z$ of a medium for a mechanical wave is defined as $Z = \rho v$, where $\rho$ is the density and $v$ is the speed of the wave in that medium. The acoustic impedance of medium M₁ is $Z_1 = \rho_1 v_1$. The acoustic impedance of medium M₂ is $Z_2 = \rho_2 v_2$.

The ratio of the transmitted amplitude to the incident amplitude is given by the formula: $\frac{A_t}{A_i} = \frac{2 Z_1}{Z_1 + Z_2}$.

Now substitute this ratio into the expression for $n$: $n = \frac{\rho_2 v_2}{\rho_1 v_1} \left(\frac{2 Z_1}{Z_1 + Z_2}\right)^2 = \frac{Z_2}{Z_1} \left(\frac{2 Z_1}{Z_1 + Z_2}\right)^2 = \frac{Z_2}{Z_1} \frac{4 Z_1^2}{(Z_1 + Z_2)^2} = \frac{4 Z_1 Z_2}{(Z_1 + Z_2)^2}$.

This formula for the intensity transmission coefficient is valid for mechanical waves, including transverse waves, at the boundary between two media.

We are given $v_2 = 0.6 v_1$, which means $v_2/v_1 = 0.6$. The relationship between the densities $\rho_1$ and $\rho_2$ is not explicitly given. However, for transverse waves in solids, the speed is $v = \sqrt{G/\rho}$, where $G$ is the shear modulus. For a wave on a string, $v = \sqrt{T/\mu}$, where $T$ is tension and $\mu$ is linear density. In many problems involving wave transmission between different media, it is implicitly assumed that the elastic property (like shear modulus $G$ or tension $T$) is the same across the boundary, while the density changes.

Assuming the elastic property is the same, i.e., $G_1 = G_2$ (for solids) or $T_1 = T_2$ (for strings). If $G_1 = G_2 = G$, then $v_1 = \sqrt{G/\rho_1}$ and $v_2 = \sqrt{G/\rho_2}$. $v_2/v_1 = \sqrt{\rho_1/\rho_2}$. We are given $v_2/v_1 = 0.6$. So, $\sqrt{\rho_1/\rho_2} = 0.6$. $\rho_1/\rho_2 = (0.6)^2 = 0.36$. $\rho_2 = \rho_1 / 0.36 = \rho_1 / (36/100) = \rho_1 \times (100/36) = \rho_1 \times (25/9)$. $\rho_2 = \frac{25}{9} \rho_1$.

Now we can calculate the impedances $Z_1$ and $Z_2$ in terms of $\rho_1$ and $v_1$: $Z_1 = \rho_1 v_1$. $Z_2 = \rho_2 v_2 = \left(\frac{25}{9} \rho_1\right) (0.6 v_1) = \frac{25}{9} \rho_1 \left(\frac{6}{10} v_1\right) = \frac{25}{9} \rho_1 \left(\frac{3}{5} v_1\right) = \frac{5 \times 3}{3 \times 3} \rho_1 v_1 = \frac{5}{3} \rho_1 v_1$. So, $Z_2 = \frac{5}{3} Z_1$.

Now substitute $Z_2 = \frac{5}{3} Z_1$ into the formula for $n$: $n = \frac{4 Z_1 Z_2}{(Z_1 + Z_2)^2} = \frac{4 Z_1 (\frac{5}{3} Z_1)}{(Z_1 + \frac{5}{3} Z_1)^2} = \frac{\frac{20}{3} Z_1^2}{(\frac{3Z_1 + 5Z_1}{3})^2} = \frac{\frac{20}{3} Z_1^2}{(\frac{8Z_1}{3})^2} = \frac{\frac{20}{3} Z_1^2}{\frac{64Z_1^2}{9}}$. $n = \frac{20}{3} \times \frac{9}{64} = \frac{20 \times 3}{64} = \frac{60}{64} = \frac{15}{16}$.

The ratio of the intensity of the transmitted and the incident waves is $n = \frac{15}{16}$. The question asks for the value of $64n$. $64n = 64 \times \frac{15}{16} = 4 \times 15 = 60$.

The assumption made was that the elastic property ($G$ or $T$) is the same in both media. This is a standard assumption for such problems unless specified otherwise. If the density were assumed to be the same, the ratio of speeds would imply a change in the elastic property, which would lead to a different result. The continuity conditions used for the amplitude ratio formula $A_t/A_i = 2Z_1/(Z_1+Z_2)$ are based on the continuity of displacement and transverse force (which depends on elastic property and displacement gradient). For this formula to be applicable with $Z = \rho v$, the force continuity requires the elastic property to be related to $\rho v^2$, i.e., $K = \rho v^2$. This is true for $v = \sqrt{K/\rho}$. Thus, assuming $K$ is constant is consistent with using $Z = \rho v$ and the derived amplitude/intensity ratios.

The final answer is $\boxed{60}$.

Explanation of the solution:

1. Write down the formula for the intensity of a wave: $I = \frac{1}{2} \rho \omega^2 A^2 v$.
2. Write the ratio of transmitted intensity to incident intensity: $n = \frac{I_t}{I_i} = \frac{\rho_2 v_2 A_t^2}{\rho_1 v_1 A_i^2}$.
3. Use the relationship between transmitted and incident amplitudes in terms of acoustic impedances $Z = \rho v$: $\frac{A_t}{A_i} = \frac{2 Z_1}{Z_1 + Z_2}$.
4. Substitute the amplitude ratio into the intensity ratio formula: $n = \frac{Z_2}{Z_1} \left(\frac{2 Z_1}{Z_1 + Z_2}\right)^2 = \frac{4 Z_1 Z_2}{(Z_1 + Z_2)^2}$.
5. Use the given speed ratio $v_2/v_1 = 0.6$. Assume the elastic property (shear modulus $G$ or tension $T$) is constant across the boundary. For transverse waves, $v = \sqrt{G/\rho}$ or $v = \sqrt{T/\mu}$. If $G$ or $T$ is constant, then $v \propto 1/\sqrt{\rho}$ or $v \propto 1/\sqrt{\mu}$.
6. From $v_2/v_1 = 0.6$, we get $\sqrt{\rho_1/\rho_2} = 0.6$, so $\rho_1/\rho_2 = 0.36 = 9/25$, which gives $\rho_2 = \frac{25}{9} \rho_1$.
7. Calculate the impedance ratio: $Z_2/Z_1 = (\rho_2 v_2) / (\rho_1 v_1) = (\frac{25}{9} \rho_1) (0.6 v_1) / (\rho_1 v_1) = \frac{25}{9} \times 0.6 = \frac{25}{9} \times \frac{3}{5} = \frac{5}{3}$. So $Z_2 = \frac{5}{3} Z_1$.
8. Substitute the impedance ratio into the formula for $n$: $n = \frac{4 Z_1 (\frac{5}{3} Z_1)}{(Z_1 + \frac{5}{3} Z_1)^2} = \frac{\frac{20}{3} Z_1^2}{(\frac{8}{3} Z_1)^2} = \frac{\frac{20}{3}}{\frac{64}{9}} = \frac{20}{3} \times \frac{9}{64} = \frac{60}{64} = \frac{15}{16}$.
9. Calculate the required value $64n = 64 \times \frac{15}{16} = 4 \times 15 = 60$.

The final answer is $\boxed{60}$.

The final answer is $\boxed{60}$.