Question

The position of a particle is given by the expression x 5 4.00 cos (3.00pt 1 p), where x is in meters and t is in seconds. Determine

(a) the frequency
(b) period of the motion
(c) the amplitude of the motion
(d) the phase constant
(e) the position of the particle at t 5 0.250 s.

Answer 1

Answer:

1.5 Hz

0.67 s

4 m

$\pi$

2.82842 m

Explanation:

The equation is

$x=4cos(3\pi t+\pi)$

It is of the form

$x=Acos(2\pi ft+\phi)$

Comparing the equations we get

$3\pi=2\pi f\\\Rightarrow f=\dfrac{3}{2}\\\Rightarrow f=1.5\ Hz$

Frequency is 1.5 Hz

Time period is given by

$T=\dfrac{1}{f}\\\Rightarrow T=\dfrac{1}{1.5}\\\Rightarrow T=0.67\ s$

The period of the motion is 0.67 s

Amplitude

$A=4\ m$

Amplitude is 4 m

Phase constant

$\phi=\pi$

The phase constant is $\pi$

At t = 0.25 s

$x=4cos(3\pi t+\pi)\\\Rightarrow x=4cos(3\pi\times 0.25+\pi)\\\Rightarrow x=2.82842\ m$

The position of the particle is 2.82842 m