Question

A particle is moving with simple harmonic motion of period 8•0 s and amplitude 5•0 m Fine (A). The speed of the particle when it's 3•0 m from the centre of it's motion (B). The maximum speed (C). The maximum acceleration

Answer 1

Answer:

a. v(1.8) = -3.14 m/s

b. $v_{max} = 3.925\ m/s$

c. $a_{max} = 3.08\ m/s^2$

Explanation:

a.

The speed of a particle executing simple harmonic motion is given by the following equation:

$v(t) = -A \omega \ Sin(\omega t)$

where,

v = velocity = ?

A = Amplitude = 5 m

ω = angular frequency = $\frac{2\pi}{Time Period} = \frac{2\pi}{8\ s}$ = 0.785 s

t = time

First we find time at 3 m by using equation for displacement:

$x = 3\ m = ACos(\omega t)\\Cos((0.785 rad/s)(t)) = \frac{3\ m}{5\ m}\\t(0.785\ rad/s) = Cos^{-1} (0.6)\\t = \frac{0.927\ rad}{0.785\ rad/s}\\t = 1.18 s$

Therefore,

$v(1.8) = velocity at 3\ m = -A\omega Sin(\omega t)\\v(1.8) = -(5\ m)(0.785\ rad/s)Sin((0.785\ rad/s)(1.18\ s))$

v(1.8) = -3.14 m/s

c.

The maximum speed is given as:

$v_{max} = A\omega\\v_{max} = (5\ m)(0.785\ rad/s)$

$v_{max} = 3.925\ m/s$

The maximum acceleration is given as:

$a_{max} = A\omega^2\\v_{max} = (5\ m)(0.785\ rad/s)^2$

$a_{max} = 3.08\ m/s^2$