High School

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------------------------------------------------ A body moving with SHM has an amplitude of 1.5m and a period of oscillation of 3 seconds. Find the velocity and acceleration of the body at t = 0.5 seconds, when time is measured from (i) the mean position and (ii) the extreme position.

Answer :

To solve this problem, we need to understand some key concepts of Simple Harmonic Motion (SHM). In SHM, a body oscillates back and forth over the same path, where its motion can be described using sinusoidal functions.


  1. Definitions and Given Values:


    • Amplitude (A): This is the maximum displacement from the mean position. Here, [tex]A = 1.5 \text{ m}[/tex].

    • Period (T): This is the time taken for one complete oscillation. Here, [tex]T = 3 \text{ seconds}[/tex].

    • Angular frequency ([tex]\omega[/tex]): This is calculated as [tex]\omega = \frac{2\pi}{T}[/tex]. So, [tex]\omega = \frac{2\pi}{3} \text{ rad/s}[/tex].



  2. Equations in SHM:


    • Displacement from the mean position (x):
      [tex]x(t) = A \cos(\omega t)[/tex]

    • Velocity (v):
      [tex]v(t) = -A\omega \sin(\omega t)[/tex]

    • Acceleration (a):
      [tex]a(t) = -A\omega^2 \cos(\omega t)[/tex]




(i) Calculations from the Mean Position at [tex]t = 0.5 \text{ seconds}[/tex]

Velocity:
[tex]v(0.5) = -1.5 \times \frac{2\pi}{3} \times \sin\left(\frac{2\pi}{3} \times 0.5\right)[/tex]
Calculating further:
[tex]v(0.5) = -1.5 \times \frac{2\pi}{3} \times \sin\left(\frac{\pi}{3}\right) = -1.5 \times \frac{2\pi}{3} \times \frac{\sqrt{3}}{2}[/tex]
[tex]v(0.5) = -\pi \sqrt{3} \approx -5.44 \text{ m/s}[/tex]

Acceleration:
[tex]a(0.5) = -1.5 \times \left(\frac{2\pi}{3}\right)^2 \times \cos\left(\frac{2\pi}{3} \times 0.5\right)[/tex]
Calculating further:
[tex]a(0.5) = -1.5 \times \left(\frac{2\pi}{3}\right)^2 \times \frac{1}{2}[/tex]
[tex]a(0.5) = -\frac{2\pi^2}{3} \approx -6.58 \text{ m/s}^2[/tex]

(ii) Calculations from the Extreme Position at [tex]t = 0.5 \text{ seconds}[/tex]

From the extreme position, the displacement uses a sine function for SHM. Here:

Velocity:
[tex]v(0.5) = A\omega \cos\left(\omega t\right)[/tex]
Calculating further:
[tex]v(0.5) = 1.5 \times \frac{2\pi}{3} \times \cos\left(\frac{2\pi}{3} \times 0.5\right) = 1.5 \times \frac{2\pi}{3} \times \frac{1}{2}[/tex]
[tex]v(0.5) = \frac{\pi}{2} \approx 1.57 \text{ m/s}[/tex]

Acceleration:
[tex]a(0.5) = -A\omega^2 \sin\left(\omega t\right)[/tex]
Calculating further:
[tex]a(0.5) = -1.5 \times \left(\frac{2\pi}{3}\right)^2 \times \sin\left(\frac{\pi}{3}\right)[/tex]
[tex]a(0.5) = -\pi^2\sqrt{3} \approx -17.08 \text{ m/s}^2[/tex]

These results give us the velocity and acceleration of the body at [tex]t = 0.5 \text{ seconds}[/tex] from both the mean and extreme positions.

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