Q112 / 150 · MHT CET 2024 (09 May Shift 2) · Physics
A musical instrument ' P ' produces sound waves of frequency ' $n$ ' and amplitude ' $A$ '. Another musical instrument ' Q ' produces sound waves of frequency $\frac{n}{4}$. The waves produced by ' $P$ ' and ' $Q$ ' have equal energies. If the amplitude of waves produced by ' P ' is ' $\mathrm{A}_{\mathrm{P}}$ ', the amplitude of waves produced by ' Q ' will be
- A$2 \mathrm{~A}_{\mathrm{P}}$
- Correct answerB$4 \mathrm{~A}_{\mathrm{P}}$
- C$6 \mathrm{~A}_{\mathrm{p}}$
- D$9 \mathrm{~A}_{\mathrm{P}}$
Solution
Energy of oscillations is given by
$E=\frac{1}{2} m \omega^2 A^2=\frac{1}{2} m(2 \pi n)^2 A^2$
$\therefore \quad E \propto \mathrm{n}^2 \mathrm{~A}^2$
As the energies are equal,
$\begin{array}{ll}
& \mathrm{n}_1^2 \mathrm{~A}_1^2=\mathrm{n}_2^2 \mathrm{~A}_2^2 \Rightarrow \mathrm{n}_1 \mathrm{~A}_1=\mathrm{n}_2 \mathrm{~A}_2 \\
\therefore \quad & \frac{\mathrm{~A}_2}{\mathrm{~A}_1}=\frac{\mathrm{n}_1}{\mathrm{n}_2}=\frac{\mathrm{n}}{\mathrm{n} / 4}=4 \\
\therefore \quad & \mathrm{~A}_2=4 \mathrm{~A}_1 \quad \\
\therefore & \left.\mathrm{~A}=4 \mathrm{~A}_{\mathrm{p}} \quad \text {... (given, } \mathrm{A}_1=A, \mathrm{~A}_2=\mathrm{A}_{\mathrm{p}}\right)
\end{array}$
Hence, the correct answer is option: 'B'.