Q155 / 180 · COMEDK 2024 (Evening Shift) · Physics
A body is executing SHM. When its displacements from the mean position are $4 \mathrm{~cm}$ and $5 \mathrm{~cm}$ it has velocity $10 \mathrm{~cms}^{-1}$ and $8 \mathrm{~cms}^{-1}$ respectively. Its periodic time $\mathrm{t}$ is
- A2$\pi$ sec
- B$\dfrac{3 \pi}{2} \mathrm{sec}$
- C$\dfrac{2 \pi}{3} \sec$
- Correct answerD$\pi$ sec
Solution
The velocity $v$ of a particle executing simple harmonic motion at a displacement $x$ is given by $v = \omega \sqrt{A^2 - x^2}$, where $A$ is the amplitude and $\omega$ is the angular frequency.
For the first case, $10 = \omega \sqrt{A^2 - 4^2} \Rightarrow 100 = \omega^2 (A^2 - 16)$.
For the second case, $8 = \omega \sqrt{A^2 - 5^2} \Rightarrow 64 = \omega^2 (A^2 - 25)$.
Dividing the two equations: $\dfrac{100}{64} = \dfrac{A^2 - 16}{A^2 - 25} \Rightarrow \dfrac{25}{16} = \dfrac{A^2 - 16}{A^2 - 25}$.
Cross-multiplying: $25(A^2 - 25) = 16(A^2 - 16) \Rightarrow 25A^2 - 625 = 16A^2 - 256$.
$9A^2 = 369 \Rightarrow A^2 = 41$.
Substituting $A^2 = 41$ into the first equation: $100 = \omega^2 (41 - 16) = 25 \omega^2$.
$\omega^2 = 4 \Rightarrow \omega = 2 \text{ rad s}^{-1}$.
The periodic time $T$ is given by $T = \dfrac{2\pi}{\omega} = \dfrac{2\pi}{2} = \pi \text{ sec}$.
Answer: $\pi$ sec
For the first case, $10 = \omega \sqrt{A^2 - 4^2} \Rightarrow 100 = \omega^2 (A^2 - 16)$.
For the second case, $8 = \omega \sqrt{A^2 - 5^2} \Rightarrow 64 = \omega^2 (A^2 - 25)$.
Dividing the two equations: $\dfrac{100}{64} = \dfrac{A^2 - 16}{A^2 - 25} \Rightarrow \dfrac{25}{16} = \dfrac{A^2 - 16}{A^2 - 25}$.
Cross-multiplying: $25(A^2 - 25) = 16(A^2 - 16) \Rightarrow 25A^2 - 625 = 16A^2 - 256$.
$9A^2 = 369 \Rightarrow A^2 = 41$.
Substituting $A^2 = 41$ into the first equation: $100 = \omega^2 (41 - 16) = 25 \omega^2$.
$\omega^2 = 4 \Rightarrow \omega = 2 \text{ rad s}^{-1}$.
The periodic time $T$ is given by $T = \dfrac{2\pi}{\omega} = \dfrac{2\pi}{2} = \pi \text{ sec}$.
Answer: $\pi$ sec
Hence, the correct answer is option: 'D'.