Animated Solution for Physics - Waves: Statement I: Two longitudinal waves given by equations— $y_1(x,t)=2a\sin (\omega t-kx)$ and $y_2(x,t)=a\sin (2\omega t-2kx)$ will have equal intensity. Statement II: Intensity of waves of given frequency in same medium is proportional to the square of amplitude only.

Imagine you are standing by a medium where two distinct longitudinal waves are propagating. The problem presents us with two statements and asks us to evaluate their truth and their logical connection.

The first statement gives us the equations of two waves:

It claims that these two waves have equal intensity. The second statement claims that the intensity of waves of a given frequency in the same medium is proportional to the square of the amplitude only. Let's break this down step by step.

To compare the intensities of these waves, we need the fundamental formula for the intensity of a longitudinal wave. Intensity is the energy transmitted per unit area per unit time, and it depends on the medium's density ($\rho$), the wave's angular frequency ($\omega$), its amplitude ($A$), and its wave velocity ($v$). The formula is:

This is a high-yield concept for JEE. Notice how intensity is highly sensitive to both the amplitude and the angular frequency—it scales with the square of both!

Let's extract the parameters for the first wave, $y_1$. By comparing it to the standard wave equation $y(x,t)=A\sin (\omega t-kx)$, we can see: - Amplitude, $A_1=2a$ - Angular frequency, ${\omega }_1=\omega$ - Wave velocity, $v_1=\frac{\omega }{k}$

Now, we substitute these values into our intensity formula:

Squaring the amplitude gives us $4a^2$. Multiplying everything out, we get:

Now, let's look at the second wave, $y_2$. Its parameters are slightly different: - Amplitude, $A_2=a$ - Angular frequency, ${\omega }_2=2\omega$ - Wave velocity, $v_2=\frac{2\omega }{2k}=\frac{\omega }{k}$

Notice that while the amplitude is halved, the angular frequency is doubled. Let's plug these into the intensity formula:

Here, squaring the angular frequency gives us $4{\omega }^2$. When we combine the factors, the $4$ from the frequency perfectly compensates for the smaller amplitude:

Look at the results! We have $I_1=\frac{2\rho {\omega }^3{a}^2}{k}$ and $I_2=\frac{2\rho {\omega }^3{a}^2}{k}$.

Since $I_1=I_2$, Statement I is absolutely true. Despite having different amplitudes and frequencies, the two waves carry the exact same energy per unit area per unit time.

Now let's evaluate Statement II. It claims that for a given frequency in the same medium, intensity is proportional to the square of the amplitude only.

Let's look back at our master formula: $I=\frac{1}{2}\rho {\omega }^2{A}^{2}v$. If the medium is the same, $\rho$ and $v$ are constant. If the frequency is 'given' (meaning it is constant), then $\omega$ is also constant. Under these specific constraints, all terms except $A^2$ become a single constant $K$.

Therefore, $I=K{A}^2$, which means $I\propto A^2$. Statement II is perfectly true.

Here is the catch where many students make a silly mistake. Does Statement II explain Statement I?

Statement II is a conditional rule—it only applies when comparing waves of the same frequency. However, the two waves in Statement I have completely different frequencies ($\omega$ and $2\omega$).

Because the condition of Statement II is violated in the scenario of Statement I, a rule that applies to a 'given frequency' cannot logically be the reason behind Statement I. Thus, both statements are true, but Statement II is not the correct explanation of Statement I.