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A 300-g object attached to the end of a spring oscillates with an amplitude of 7.0 cm and a frequency of 1.80 Hz. (a) Find its maximum speed and maximum acceleration. (b) What is its speed when it is 3.0 cm from its equilibrium position

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Answer:

a) v = 0.79 m / s, a = 8.95 m / s², b) v = -0.3285 m / s

Step-by-step explanation:

A simple harmonic motion is described by the expression

x = A cos (wt + Ф)

the range of motion is A = 7.0 cm = 0.070 m

angular velocity and frequency are related

w = 2πf

w = 2π 1.80

w = 11.3 rad / s

we substitute

x = 0.070 cos (11.3t +Ф)

a) to find the velocity we use

v = dx / dt

v = - Aw sin (wt + Ф)

the maximum velocity when the cosine argument is π/2 0 3π/2 therefore the sine function is ±1

v = A w

v = 0.070 11.3

v = 0.79 m / s

the acceleration is

a = dv / dt

a = - A w² cos (wt + Ф)

the acceleration is maximum for an angle of o or pi, consequently the cosine works worth ±1

a = A w²

a = 0.070 11.3²

a = 8.95 m / s²

b) Let's find the time it takes to get to x = 0.03 m

wt + Ф = x / A

wt + Ф = 0.03 / 0.07 = 0.42857

To find the value of fi the initial conditions are used, in general if the system is released from rest fi = 0

t = 0.42857 / w

t = 0.42857 / 11.3

t = 0.0379 s

the speed for this time is

v = -A w sin wt

v = - 0.07 11.3 sin (11.3 0.0379)

remember angles are in radians

v = -0.3285 m / s

User Damian Winnicki
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