Q122 / 180 · COMEDK 2025 (25 May Shift) · Physics
From the equations given below select the correct equation representing the displacement of a particle undergoing a periodic motion. [$y$ - displacement; a - amplitude $T$ - Period; $\omega$ - angular velocity; $t$-time]
- A$y=a \sin \left[\dfrac{\omega \mathrm{t}}{T}\right]$
- Correct answerB$y=a \sin \left[\dfrac{(2 \pi \mathrm{t})}{T}\right]$
- C$y=a \sin \left[\dfrac{(2 \pi \omega)}{T}\right]$
- D$y=a \omega \sin \left[\dfrac{(2 \pi \mathrm{t})}{T}\right]$
Solution
The general equation for the displacement of a particle undergoing simple harmonic motion is given by $y = a \sin(\omega t + \phi)$, where $\omega$ is the angular frequency and $t$ is the time.
The relationship between angular frequency $\omega$ and the time period $T$ is defined as $\omega = \dfrac{2 \pi}{T}$.
Substituting this expression for $\omega$ into the general displacement equation, we get $y = a \sin\left(\dfrac{2 \pi}{T} t\right)$.
Rearranging the terms inside the sine function, we obtain $y = a \sin\left(\dfrac{2 \pi t}{T}\right)$.
Answer: $y=a \sin \left[\dfrac{(2 \pi \mathrm{t})}{T}\right]$
The relationship between angular frequency $\omega$ and the time period $T$ is defined as $\omega = \dfrac{2 \pi}{T}$.
Substituting this expression for $\omega$ into the general displacement equation, we get $y = a \sin\left(\dfrac{2 \pi}{T} t\right)$.
Rearranging the terms inside the sine function, we obtain $y = a \sin\left(\dfrac{2 \pi t}{T}\right)$.
Answer: $y=a \sin \left[\dfrac{(2 \pi \mathrm{t})}{T}\right]$
Hence, the correct answer is option: 'B'.