Energy and Intensity of Waves Questions in English

(A) For a point source,the intensity $I$ at a distance $r$ is given by the inverse square law: $I \propto \frac{1}{r^2}$.
Let $I_0$ be the intensity at $r_0 = 1 \text{ m}$,where $I_0 = 16 \times 10^{-8} \text{ W m}^{-2}$.
The intensity at distance $r$ is $I(r) = \frac{I_0 \cdot r_0^2}{r^2} = \frac{16 \times 10^{-8} \times (1)^2}{r^2} = \frac{16 \times 10^{-8}}{r^2} \text{ W m}^{-2}$.
At $r_1 = 2 \text{ m}$,the intensity is $I_1 = \frac{16 \times 10^{-8}}{2^2} = \frac{16 \times 10^{-8}}{4} = 4 \times 10^{-8} \text{ W m}^{-2}$.
At $r_2 = 4 \text{ m}$,the intensity is $I_2 = \frac{16 \times 10^{-8}}{4^2} = \frac{16 \times 10^{-8}}{16} = 1 \times 10^{-8} \text{ W m}^{-2}$.
The difference in intensity is $|I_1 - I_2| = |4 \times 10^{-8} - 1 \times 10^{-8}| = 3 \times 10^{-8} \text{ W m}^{-2}$.
Thus,the value is $3$.

(A) The energy of oscillations is given by the formula $E = \frac{1}{2} m \omega^2 A^2$.
Since $\omega = 2 \pi n$,we have $E \propto n^2 A^2$.
Given that the energies of the waves produced by $X$ and $Y$ are equal,we can write:
$n_X^2 A_X^2 = n_Y^2 A_Y^2$.
Substituting the given values $n_X = n$,$A_X = A$,and $n_Y = \frac{n}{3}$:
$n^2 A^2 = (\frac{n}{3})^2 A_Y^2$.
$n^2 A^2 = \frac{n^2}{9} A_Y^2$.
$A^2 = \frac{A_Y^2}{9}$.
$A_Y^2 = 9 A^2$.
Taking the square root on both sides,we get $A_Y = 3 A$.

(B) The energy $E$ of a sound wave is proportional to the square of its amplitude $A$ and the square of its frequency $n$. Mathematically,$E \propto A^2 n^2$.
Since the energies of the waves produced by $P$ and $Q$ are equal,we have $E_P = E_Q$.
This implies $A_P^2 n_P^2 = A_Q^2 n_Q^2$.
Given $n_P = n$ and $n_Q = \frac{n}{4}$,we substitute these values into the equation:
$A_P^2 n^2 = A_Q^2 (\frac{n}{4})^2$.
$A_P^2 n^2 = A_Q^2 (\frac{n^2}{16})$.
Dividing both sides by $n^2$,we get $A_P^2 = \frac{A_Q^2}{16}$.
$A_Q^2 = 16 A_P^2$.
Taking the square root of both sides,we get $A_Q = 4 A_P$.

(D) For a spherical wave emitted by a point source,the intensity $I$ at a distance $r$ is given by $I = \frac{P}{4 \pi r^2}$,where $P$ is the power of the source.
Since $I \propto A^2$,where $A$ is the amplitude,we have $A^2 \propto \frac{1}{r^2}$,which implies $A \propto \frac{1}{r}$.
Let $A_P$ and $A_Q$ be the amplitudes at distances $r_P = 4 \ m$ and $r_Q = 9 \ m$ respectively.
Then,the ratio of amplitudes is $\frac{A_P}{A_Q} = \frac{r_Q}{r_P}$.
Substituting the given values,$\frac{A_P}{A_Q} = \frac{9}{4}$.

(B) The intensity $(I)$ of a sound wave is directly proportional to the square of its amplitude $(A)$.
$I \propto A^2$
Therefore,the ratio of intensities is related to the ratio of amplitudes as:
$\frac{I_{\text{reflected}}}{I_{\text{incident}}} = \left( \frac{A_{\text{reflected}}}{A_{\text{incident}}} \right)^2$
Given that the intensity decreases by $20 \%$,the reflected intensity is:
$I_{\text{reflected}} = I_{\text{incident}} - 0.20 I_{\text{incident}} = 0.8 I_{\text{incident}} = \frac{4}{5} I_{\text{incident}}$
Substituting this into the ratio equation:
$\frac{\frac{4}{5} I_{\text{incident}}}{I_{\text{incident}}} = \left( \frac{A_{\text{reflected}}}{A} \right)^2$
$\frac{4}{5} = \left( \frac{A_{\text{reflected}}}{A} \right)^2$
Taking the square root on both sides:
$A_{\text{reflected}} = \sqrt{\frac{4}{5}} A = \frac{2}{\sqrt{5}} A$

(D) The volume of a sphere is given by $V = \frac{4}{3} \pi R^3$,which implies $V \propto R^3$.
If the volume $V$ increases by $8$ times,then $R^3$ increases by $8$ times,so the radius $R$ increases by $2$ times $(R \rightarrow 2R)$.
The surface area of the sphere is $A = 4 \pi R^2$,which implies $A \propto R^2$.
Since $R$ increases by $2$ times,the area $A$ increases by $2^2 = 4$ times $(A \rightarrow 4A)$.
Intensity $I$ is defined as power per unit area,$I = \frac{P}{A}$.
Since the power $P$ of the point source remains constant,$I \propto \frac{1}{A}$.
Therefore,if the area $A$ increases by $4$ times,the intensity $I$ will decrease by $4$ times $(I \rightarrow \frac{I}{4})$.