An object with mass 3.0 kg is executing simple harmonic motion, attached to a spring with spring constant k =210 N/m. When the object is 0.23 m from its equilibrium position, it is moving with a speed of 0.72 m/s.

(a) Calculate the amplitude of the motion.
m

(b) Calculate the maximum speed attained by the object.
m/s

F = m a = -kx

if x = A cos w t
v = dx/dt = -Aw sin w t
a = d^2x/dt^2 = -A w^2 cos w t = -w^2 x

so
-m w^2 x = -k x
w^2 = k/m
w^2 = 210/3 = 70
w = sqrt(70) = 8.37

when |x| = .23 , |v| = .72
.23 = A cos 8.37 t
.72 = -A (8.37) sin 8.37 t
so
cos 8.37 t = .23/A
sin 8.37 t = .086/A

cos^2 + sin^2 = 1
.0529 + .007399 = A^2
A^2 = .060299
A = .246 m
Vmax = Aw = .246(8.37) = 2.06 m/s

check my arithmetic !!!

To solve this problem, we can use the equations related to simple harmonic motion:

(a) The amplitude of simple harmonic motion can be calculated using the equation:

Amplitude (A) = x_max - x_eq

Where:
- x_max is the maximum displacement from the equilibrium position
- x_eq is the equilibrium position

Given that the object is 0.23 m from its equilibrium position, we know that x_max = 0.23 m.
Since the object is executing simple harmonic motion, the equilibrium position is at x = 0. Therefore, x_eq = 0.

Substituting these values into the equation, we can calculate the amplitude:

Amplitude = 0.23 m - 0 m = 0.23 m

Therefore, the amplitude of the motion is 0.23 m.

(b) The maximum speed attained by the object occurs when it is at the equilibrium position. At this point, all the potential energy is converted into kinetic energy, resulting in maximum speed.

The potential energy stored in a spring is given by the equation:

Potential energy (PE) = (1/2) * k * A^2

Where:
- k is the spring constant
- A is the amplitude

For maximum speed, the potential energy is converted entirely into kinetic energy:

Potential energy (PE) = Kinetic energy (KE)

So, we can set up the equation:

(1/2) * k * A^2 = (1/2) * m * v_max^2

Given that the mass (m) of the object is 3.0 kg and the spring constant (k) is 210 N/m, we can solve for the maximum speed (v_max).

Rearranging the equation:

v_max^2 = (k * A^2) / m

Substituting the given values:

v_max^2 = (210 N/m * (0.23 m)^2) / 3.0 kg

Calculating the right-hand side of the equation:

v_max^2 = 0.017055 N * m / kg

Taking the square root to find v_max:

v_max = √(0.017055 N * m / kg)

v_max = 0.1307 m/s (rounded to four decimal places)

Therefore, the maximum speed attained by the object is 0.1307 m/s.