Question

A particle executes simple harmonic motion with an amplitude of $10cm$ and time period $6s$ . At $t = 0$ it is at position $x = 5cm$ from mean position and going towards positive x-direction. Write the equation for the displacement $x$ at time $t$ . Find the magnitude of the acceleration of the particle at $t = 4s$ .

Answer 1

Hint : In order to obtain the equation of displacement of $x$ with respect to the time $t$ , we have to first find the angular frequency of the simple harmonic motion. Then by using this equation of displacement, we can calculate the magnitude of acceleration of the particle at $t = 4s$ .
$\omega = \dfrac{{2\pi }}{T} \\\ Y = A\sin (\omega t + \phi ) \\\ a = - {\omega ^2}x \\\$

Complete Step By Step Answer:
From the question, we know that the amplitude of simple harmonic motion, $A = 10cm$ and the time period of the simple harmonic motion, $T = 6s$ .
When the time is at $t = 0$ , the particle is at a position $x = 5cm$ from the mean position.
In order to obtain the equation of displacement of $x$ with respect to the time $t$ , we have to first find the angular frequency of the simple harmonic motion.
Angular frequency $(\omega )$ is given by,
$\omega = \dfrac{{2\pi }}{T} = \dfrac{{2\pi }}{6} = \dfrac{\pi }{3}{\sec ^{ - 1}}$
We have obtained the angular frequency which is $\omega = \dfrac{\pi }{3}{\sec ^{ - 1}}$
Now, let’s consider the equation of simple harmonic motion which is given by,
$Y = A\sin (\omega t + \phi )$ [let this be equation (1)]
where $Y$ is the displacement of the particle, and $\phi$ is the phase of the particle.
Now, we shall substitute the values of $A,t$ and $\omega$ in equation (1).
Then we get,
$5 = 10\sin (\omega \times 0 + \phi ) \\\ \Rightarrow 5 = 10\sin \phi \\\ \Rightarrow \sin \phi = \dfrac{1}{2} \\\ \Rightarrow \phi = \dfrac{\pi }{6} \\\$
Therefore, the equation of displacement can be written as,

$x = (10cm)\sin (\dfrac{\pi }{3}t + \dfrac{\pi }{6})$
Now, we have to find the magnitude of the acceleration of the particle at $t = 4s$ . So, we substitute the value $t = 4$ in the equation of displacement.
$x = (10)\sin (\dfrac{\pi }{3}(4) + \dfrac{\pi }{6})$
By taking LCM, we get,
$x = 10\sin (\dfrac{{8\pi + \pi }}{6}) \\\ = 10\sin (\dfrac{{9\pi }}{6}) \\\ = 10\sin (\dfrac{{3\pi }}{2}) \\\ = 10\sin (\pi + \dfrac{\pi }{2}) \\\$
Since $\sin (\pi + \phi ) = - \sin (\phi )$
We get,
$x = - 10\sin (\dfrac{\pi }{2})$
Since $\sin (\dfrac{\pi }{2}) = 1$
$x = - 10$
Therefore, we found that the displacement of the particle, $x = - 10cm$
The magnitude of acceleration of the particle is given by the formula,
$a = - {\omega ^2}x$
Now, we substitute the values of $\omega$ and $x$ in the above equation and we get,
$a = - {(\dfrac{\pi }{3})^2} \times ( - 10) \\\ = (\dfrac{{{\pi ^2}}}{9}) \times 10 \\\ = 10.95 \\\ \approx 11 \\\$
Therefore, the magnitude of acceleration of the particle is $a \approx 11cm.{\sec ^{ - 2}}$ .

Note :
If the magnitude of acceleration obtained is a negative value, this indicates that the acceleration and the displacement are in the opposite direction of each other. If the magnitude of acceleration obtained is a positive value, this indicates that the acceleration and the displacement are in the same direction. So, the students have to be careful while solving, as a negative sign makes a lot of difference.