Question

Determine the maximum velocity and maximum acceleration of a particle which moves in simple harmonic motion with an amplitude of 0.2 in and a period of 0.1 s.

Answer 1

Answer:

The maximum velocity and maximum acceleration of the particle are approximately 12.566 inches per second and 789.568 inches per square second.

Explanation:

The equation of motion for the position of a particle experimentating a simple harmonic motion ($x(t)$), measured in inches, is described by the following expression:

$x(t) = A\cdot \cos \left(\frac{2\pi\cdot t}{T} +\phi\right)$ (1)

Where:

$A$ - Amplitude, measured in inches.

$t$ - Time, measured in seconds.

$T$ - Period, measured in seconds.

$\phi$ - Phase, measured in radians.

Then, we obtain the formulas for the velocity and acceleration of the particle by differentiating (1):

$v(t) = -\frac{2\pi\cdot A}{T}\cdot \sin \left(\frac{2\pi\cdot t}{T}+\phi \right)$ (2)

$a(t) = -\left(\frac{2\pi}{T} \right)^{2}\cdot A\cdot \cos \left(\frac{2\pi\cdot t}{T}+\phi \right)$ (3)

From (2) and (3) we find that maximum velocity ($v_{max}$), measured in inches per second, and maximum acceleration ($a_{max}$), measured in inches per square second, are defined by the following formulas:

$v_{max} = \frac{2\pi\cdot A}{T}$ (4)

$a_{max} = \left(\frac{2\pi}{T} \right)^{2}\cdot A$ (5)

If we know that $A = 0.2\,in$ and $T = 0.1\,s$, then the maximum velocity and maximum acceleration of the particle are, respectively:

$v_{max} = \frac{2\pi\cdot (0.2\,in)}{0.1\,s}$

$v_{max} \approx 12.566\,\frac{in}{s}$

$a_{max} = \left(\frac{2\pi}{0.1\,s} \right)^{2}\cdot (0.2\,in)$

$a_{max} \approx 789.568\,\frac{in}{s^{2}}$

The maximum velocity and maximum acceleration of the particle are approximately 12.566 inches per second and 789.568 inches per square second.