Question

If the object-spring system is described by x = (0.340 m) cos (1.55t), find the following.

(a) the amplitude, the angular frequency, the frequency, and the period

A =	m
ω =	rad/s
f =	Hz
T =	s

(b) the maximum magnitudes of the velocity and the acceleration

vmax =	m/s
amax =	m/s2

(c) the position, velocity, and acceleration when t = 0.250 s

x =	m
v =	m/s
a =	m/s2

Answer 1

Displacement of a spring in simple harmonic motion is given by,
x(t) = Acos(t + )

where x is the position
A is the amplitude
is the angular frequency
is the  phase

For this problem,

x = (0.340 m) cos (1.55t)
Just find the corresponding numbers.
A = 0.340 m
= 1.55 rad/s

The frequency is given by
f = /2
f = (1.55 rad/s) / 2
f = 0.246 Hz

The period is just the inverse of the frequency.
T = 1/f
T = 1/(0.246 Hz)
T = 4.05 s
B)Next we need to find the velocity and acceleration,
v(t) = -Acos(t + )
and
a(t) = A²cos(t +)

Then maximum velocity is A, and the maximum acceleration is A².

The maximum velocity is v_max
v_max = A = (0.340 m)(1.55 rad/s) = 0.527 m/s
a_max= A² =(0.340 m)(1.55 rad/s)² =0.82 m/s²
c) Plug in the value for t, let us assume the phase =0.
x(t) = Acos(t + )
x(0.250 s) = (0.340 m)cos((1.55 rad/s)(0.250 s) + 0 rad)
x(0.250 s) = 0.339 m

v(t) = -Acos(t + )
v(0.250 s) = -(0.340 m)(1.55 rad/s)cos((1.55 rad/s)(0.250 s) + 0 rad)
v(0.250 s) = -0.526 m/s
a(t) = A²cos(t + )
a(0.250 s) = (0.340 m)(1.55 rad/s)²cos((1.55 rad/s)(0.250 s) + 0 rad)
a(0.250 s) = 0.817 m/s²