A particle executes simple harmonic motion with an amplitude of 10 cm and a time period of 6 s . At t = 0 , it is located at x = 5 cm and is moving towards the positive x-direction. Determine the equation for the

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Quantrex Academy · Accepted Answer

Correct answer: D. x = 10 ((π)/(3)t + (π)/(6)) cm , 11 cm/s^2

The angular frequency of the particle is

\omega = \dfrac{2\pi}{T} = \dfrac{2\pi}{6} = \dfrac{\pi}{3} \text{ rad/s}

.

The general equation for displacement in SHM is

x(t) = A \sin(\omega t + \phi_0)

.

Given

A = 10 \text{ cm}

, and at

t = 0

,

x = 5 \text{ cm}

:

10 \sin(\phi_0) = 5 \Rightarrow \sin(\phi_0) = \dfrac{1}{2} \Rightarrow \phi_0 = \dfrac{\pi}{6}

or

\dfrac{5\pi}{6}

.

The velocity is

v(t) = A\omega \cos(\omega t + \phi_0)

. Since the particle is moving towards the positive x-direction at

t = 0

,

v(0) > 0

, which implies

\cos(\phi_0) > 0

.

Therefore,

\phi_0 = \dfrac{\pi}{6}

.

The equation for displacement is

x(t) = 10 \sin\left(\dfrac{\pi}{3}t + \dfrac{\pi}{6}\right) \text{ cm}

.

At

t = 4 \text{ s}

, the displacement is:

x(4) = 10 \sin\left(\dfrac{4\pi}{3} + \dfrac{\pi}{6}\right) = 10 \sin\left(\dfrac{9\pi}{6}\right) = 10 \sin\left(\dfrac{3\pi}{2}\right) = -10 \text{ cm}

.

The magnitude of acceleration is

|a| = \omega^2 |x|

:

|a| = \left(\dfrac{\pi}{3}\right)^2 \times 10 = \dfrac{10\pi^2}{9}

Taking

\pi^2 \approx 9.87

, we get

|a| \approx \dfrac{98.7}{9} \approx 11 \text{ cm/s}^2

.

A particle executes simple harmonic motion with an amplitude of 10...

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