Answer: Given that A 2.75 kg particle moves as function of time as follows: x(t) = 5cos(1.25t+π/4)
a.) An Amplitude is the maximum displacement from the original position.
The amplitude is 5 meters
The angular frequency (w) = 2πF
The Angular frequency = 1.25 rad/s
We can calculate frequency from angular frequency formula
w = 2πF
1.25 = 2 x 22/7 x f
F = 1.25/6.283
F = 0.199 Hz
F = 0.2 Hz (Approximately)
Period is a reciprocal of frequency
Period T = 1/F = 1/0.2
Period = 5 second
b.) To find the equation of the velocity, you will differentiate
x(t) = 5cos(1.25t+π/4)
Therefore, Velocity V = -6.25sin(1.25t+π/4)
c.) To find the equation of the acceleration of the particle, differentiate velocity
Acceleration a = 7.8125cos(1.25t+π/4)
d.) The spring constant K = Force/extension
Force = ma and Extension = distance x
K = 2.75 x 7.8125cos(1.25t+π/4) / 5cos(1.25t+π/4)
k = 21.4843/5
K = 4.3
e.) K.E = 1/2mv^2
K.E = 0.5 x 2.75 x [-6.25sin(1.25t+π/4)]^2
P.E = mgh
P.E = 1/2 KA^2
f.) Total Energy = P.E + K.E
= 1/2KA^2 + 1/2mv^2
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