An object with mass 3.0 kg is executing simple harmonic motion, attached to a spring with spring constant 250 N/m .

Question

When the object is 0.018 m from its equilibrium position, it is moving with a speed of 0.65 m/s :

Calculate the amplitude of the motion AND Calculate the maximum speed attained by the object.

Answer 1

ω = √(k/m) = √(250/3)

The position is found by x(t) = Acos(ωt)

So 0.018 = Acos(√(250/3)t)

The velocity is found by v(t) = -Aωsin(ωt)

So 0.65 = -A(√(250/3))sin(√(250/3)t)

Let's divide the velocity equation by the position equation:

0.65/0.018 = [-A(√(250/3))sin(√(250/3)t)]/[Acos(√(250/3)t)]

36.1111 = -√(250/3)tan(√(250/3)t)

-3.955774 = tan(√(250/3)t)

arctan(-3.955774) = √(250/3)t

-1.3231888 = √(250/3)t

t = -0.14495

Using that time, and plugging it into 0.018 = Acos(√(250/3)t) we get:

0.018 = Acos(√(250/3)( -0.14495))

A = 0.018/cos(√(250/3)( -0.14495))

A = 0.018/0.245085

A = 0.073 m

The motion is modeled then by:

x(t) = 0.073cos(√(250/3)t)

The velocity is modeled by:

v(t) = -Aωsin(ωt)

v(t) = -(0.073)(√250/3)sin(√(250/3)t)

v(t) = -0.670448sin(√(250/3)t)

Since sin(whatever) oscillates between -1 and 1, the largest velocity occurs when sin() = -1 making the maximum velocity (-0.670448)(-1) = 0.67 m/s