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2 years ago

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The following represents a mass attached to a spring oscillating in simple harmonic motion. X(t) = 4.0 cos(3.0t +0.10) units of

meters, radians and seconds
a. What is the amplitude? (1)
b. What is the angular frequency? (1)
c. What is the frequency? (2)
d. What is the period? (2)
e. What is the phase constant? (1)
f. What is the maximum speed? (2)
g. If the mass m= 1.2 kg, what is the spring constant? (2)
h. If the mass m= 1.2 kg, what is the total energy of the oscillator? (3)
i. What is the potential energy of the oscillator at t=0 s? (3)
j. What is the kinetic energy of the oscillator at t=0 s? (3)

1 answer:

kolbaska11 [484]2 years ago

3 0

Answer:

a) A = 4.0 m , b) w = 3.0 rad / s , c) f = 0.477 Hz , d) T = 20.94 s

Explanation:

The equation that describes the oscillatory motion is

x = A cos (wt + fi)

In the exercise we are told that the expression is

x = 4.0 cos (3.0 t + 0.10)

let's answer the different questions

a) the amplitude is

A = 4.0 m

b) the frequency or angular velocity

w = 3.0 rad / s

c) angular velocity and frequency are related

w = 2π f

f = w / 2π

f = 3 / 2π

f = 0.477 Hz

d) the period

frequency and period are related

T = 1 / f

T = 1 / 0.477

T = 20.94 s

e) the phase constant

Ф = 0.10 rad

f) velocity is defined by

v = dx / dt

v = - A w sin (wt + Ф)

speed is maximum when sine is + -1

v = A w

v = 4 3

v = 12 m / s

g) the angular velocity is

w² = k / m

k = m w²

k = 1.2 3²

k = 10.8 N / m

h) the total energy of the oscillator is

Em = ½ k A²

Em = ½ 10.8 4²

Em = 43.2 J

i) the potential energy is

Ke = ½ k x²

for t = 0 x = 4 cos (0 + 0.1)

x = 3.98 m

j) kinetic energy

K = ½ m v²

for t = 00.1 ²

v = A w sin 0.10

v = 4 3 sin 0.10

v = 1.98 m / s

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