A simple harmonic motion of a particle is represented as, y = 10 cos ω

A simple harmonic motion of a particle is represented as, y = 10 cos ωt 10. The acceleration of the particle at time t = $\frac{\pi}{2\omega}$ will be : (symbols here carry their usual meanings)

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10 ω

$-10omega^2$

$ rac{10}{omega}$

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UPSC CAPF – 2023

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The correct answer is 0.

The given equation for the simple harmonic motion of a particle is $y = 10 \cos \omega t + 10$. This represents the displacement of the particle from a reference point (in this case, an origin shifted by 10 units). The velocity of the particle is the first derivative of displacement with respect to time: $v = \frac{dy}{dt} = \frac{d}{dt}(10 \cos \omega t + 10) = -10 \omega \sin \omega t$. The acceleration of the particle is the first derivative of velocity with respect to time: $a = \frac{dv}{dt} = \frac{d}{dt}(-10 \omega \sin \omega t) = -10 \omega^2 \cos \omega t$. We need to find the acceleration at time $t = \frac{\pi}{2\omega}$. Substituting this value of $t$ into the acceleration equation: $a\left(t=\frac{\pi}{2\omega}\right) = -10 \omega^2 \cos\left(\omega \cdot \frac{\pi}{2\omega}\right) = -10 \omega^2 \cos\left(\frac{\pi}{2}\right)$. Since $\cos\left(\frac{\pi}{2}\right) = 0$, the acceleration at this time is $a = -10 \omega^2 \cdot 0 = 0$.

In simple harmonic motion described by $y = A \cos(\omega t + \phi) + C$, the term $A \cos(\omega t + \phi)$ represents the oscillation about the equilibrium position. The acceleration is proportional to the displacement from the equilibrium position and directed towards it ($a = -\omega^2 (y-C)$). In this case, the equilibrium position is at $y=10$. At $t = \frac{\pi}{2\omega}$, the displacement $y = 10 \cos(\frac{\pi}{2}) + 10 = 10(0) + 10 = 10$. Since the displacement is equal to the equilibrium position, the acceleration is zero, as expected.

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