Question

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

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

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

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 vmax
vmax = A = (0.340 m)(1.55 rad/s) = 0.527 m/s
amax= 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) = 0.339 m

v(t) = -Acos(t + )
v(0.250 s) = -0.526 m/s
a(t) = A2cos(t + )
a(0.250 s) = 0.817 m/s2

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