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

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

b.
`v_(max) = 3.925\ m/s`

c.
`a_(max) = 3.08\ m/s^2`

Step-by-step 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 =
`(2\pi)/(Time Period) = (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)) = (3\ m)/(5\ m)\\t(0.785\ rad/s) = Cos^(-1) (0.6)\\t = (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`