Q128 / 180 · COMEDK 2025 (25 May Shift) · Physics
A particle is executing simple harmonic motion. The displacement of the particle from the mean position in 2 second is equal to $\dfrac{1}{\sqrt{2}}$ times its amplitude. What is the period of oscillation of the particle ?
- Correct answerA$\mathrm{T}=16 \mathrm{~s}$
- B$\mathrm{T}=4 \mathrm{~s}$
- C$\mathrm{T}=2 \mathrm{~s}$
- D$\mathrm{T}=8 \mathrm{~s}$
Solution
The displacement of a particle in simple harmonic motion is given by $x(t) = A \sin(\omega t + \phi)$. Assuming the particle starts from the mean position at $t = 0$, we have $\phi = 0$, so $x(t) = A \sin(\omega t)$.
Given that at $t = 2$ s, $x(2) = \dfrac{A}{\sqrt{2}}$, we substitute these values into the equation:
$\dfrac{A}{\sqrt{2}} = A \sin(2\omega)$
$\sin(2\omega) = \dfrac{1}{\sqrt{2}}$
This implies $2\omega = \dfrac{\pi}{4}$ or $2\omega = \dfrac{3\pi}{4}$. Taking the simplest case for the first occurrence, $2\omega = \dfrac{\pi}{4}$.
Since $\omega = \dfrac{2\pi}{T}$, we have $2 \left( \dfrac{2\pi}{T} \right) = \dfrac{\pi}{4}$ or $2 \left( \dfrac{2\pi}{T} \right) = \dfrac{3\pi}{4}$.
For $2\omega = \dfrac{\pi}{4}$, $\dfrac{4\pi}{T} = \dfrac{\pi}{4} \Rightarrow T = 16$ s.
For $2\omega = \dfrac{3\pi}{4}$, $\dfrac{4\pi}{T} = \dfrac{3\pi}{4} \Rightarrow T = \dfrac{16}{3}$ s.
Comparing with the given options, $T = 16$ s is the correct period.
Answer: $\mathrm{T}=16 \mathrm{~s}$
Given that at $t = 2$ s, $x(2) = \dfrac{A}{\sqrt{2}}$, we substitute these values into the equation:
$\dfrac{A}{\sqrt{2}} = A \sin(2\omega)$
$\sin(2\omega) = \dfrac{1}{\sqrt{2}}$
This implies $2\omega = \dfrac{\pi}{4}$ or $2\omega = \dfrac{3\pi}{4}$. Taking the simplest case for the first occurrence, $2\omega = \dfrac{\pi}{4}$.
Since $\omega = \dfrac{2\pi}{T}$, we have $2 \left( \dfrac{2\pi}{T} \right) = \dfrac{\pi}{4}$ or $2 \left( \dfrac{2\pi}{T} \right) = \dfrac{3\pi}{4}$.
For $2\omega = \dfrac{\pi}{4}$, $\dfrac{4\pi}{T} = \dfrac{\pi}{4} \Rightarrow T = 16$ s.
For $2\omega = \dfrac{3\pi}{4}$, $\dfrac{4\pi}{T} = \dfrac{3\pi}{4} \Rightarrow T = \dfrac{16}{3}$ s.
Comparing with the given options, $T = 16$ s is the correct period.
Answer: $\mathrm{T}=16 \mathrm{~s}$
Hence, the correct answer is option: 'A'.