A 0.46 kg object connected to a light spring with a spring constant of 18.0 N/m oscillates on a frictionless horizontal surface. The spring is compressed 4.0 cm and released from rest.

(a) Determine the maximum speed of the mass.
unit = cm/s
(b) Determine the speed of the mass when the spring is compressed 1.5 cm.
unit = cm/s
(c) Determine the speed of the mass when the spring is stretched 1.5 cm.
unit = cm/s
(d) For what value of x does the speed equal one-half the maximum speed?
unit = cm

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To find the answers to these questions, we'll need to use the equations of motion for simple harmonic motion.

First, let's find the angular frequency, ω, of the oscillation. The angular frequency is given by the formula ω = √(k / m), where k is the spring constant and m is the mass of the object. In this case, k = 18.0 N/m and m = 0.46 kg.

(a) To determine the maximum speed of the mass, we use the equation vmax = Aω, where A is the amplitude of the oscillation (the maximum displacement from equilibrium). In this case, the spring is compressed 4.0 cm, so the amplitude is 4.0 cm = 0.04 m.

Substituting the values into the equation, we have vmax = (0.04 m) * √(18.0 N/m / 0.46 kg).

To convert the result to cm/s, we multiply the value by 100, as 1 m equals 100 cm.

(b) To determine the speed of the mass when the spring is compressed 1.5 cm, we can use the same equation vmax = Aω, but with a different amplitude A. In this case, the amplitude is 1.5 cm = 0.015 m.

Substituting the values into the equation, we have vmax = (0.015 m) * √(18.0 N/m / 0.46 kg).

To convert the result to cm/s, we multiply the value by 100.

(c) To determine the speed of the mass when the spring is stretched 1.5 cm, we can again use the same equation, but with a negative amplitude A (since the spring is now stretched in the opposite direction). In this case, the amplitude is -1.5 cm = -0.015 m.

Substituting the values into the equation, we have vmax = (-0.015 m) * √(18.0 N/m / 0.46 kg).

To convert the result to cm/s, we multiply the value by 100.

(d) To find the value of x for which the speed is one-half the maximum speed, we recall that the speed of the mass is given by v = ω * sqrt(A^2 - x^2), where x is the displacement from equilibrium. We want to find the value of x when v = vmax/2.

Substituting the values into the equation, we have vmax/2 = √(18.0 N/m / 0.46 kg) * √(0.04 m^2 - x^2).

Squaring both sides of the equation, we get (vmax/2)^2 = (18.0 N/m / 0.46 kg) * (0.04 m^2 - x^2).

Rearranging the equation to isolate x^2, we have x^2 = 0.04 m^2 - (vmax/2)^2 * (0.46 kg / 18.0 N/m).

Taking the square root of both sides, we get x = √(0.04 m^2 - (vmax/2)^2 * (0.46 kg / 18.0 N/m)).

To convert the result to cm, we multiply the value by 100.