A 5.0-kg object moves on a horizontal frictionless surface under the influence of a spring with
force constant 10
3
N/m. The object is displaced 50 cm and given an initial velocity of 10 m/s back
toward the equilibrium position. (a) What is the velocity of the motion? (b) How much energy is
associated with the motion? (c) What is the amplitude of the oscillation?
Your school name is not the subject.
Is the force constant of the spring 10^3 N/m?
(a). Did you omit a word? The velocity of the motion changes during oscillation.
To find the answers to these questions, we need to use the principles of spring-mass systems and the conservation of mechanical energy.
(a) The velocity of the motion can be found using the equation for the motion of a mass-spring system:
v = -ωA sin(ωt + φ)
Where:
v = velocity of the mass
A = amplitude of the oscillation
ω = angular frequency = √(k/m), where k is the force constant of the spring and m is the mass of the object
In this case, the object has an initial velocity of 10 m/s back towards the equilibrium position. Since the amplitude is given to be 50 cm, which is equivalent to 0.5 m, and the mass is 5.0 kg, we can calculate the angular frequency:
ω = √(k / m) = √(1000 N/m / 5.0 kg) = √(200) ≈ 14.14 rad/s
Now, substituting the values into the equation for velocity:
v = - (14.14 rad/s) (0.5 m) sin(14.14 t + φ)
To determine the initial phase angle (φ), we use the information that the object is at its maximum amplitude (50 cm) when t = 0. Therefore, sin(φ) = 1, which means φ = π/2.
Replacing φ, we get:
v = - (14.14 rad/s) (0.5 m) sin(14.14 t + π/2)
v = - 7.07 m/s sin(14.14 t + π/2)
So, the velocity of the motion is given by v = - 7.07 m/s sin(14.14 t + π/2).
(b) The energy associated with the motion can be calculated using the equation for the total mechanical energy of a mass-spring system:
E = (1/2) kA^2
Where:
E = total mechanical energy
A = amplitude of the oscillation
k = force constant of the spring
Substituting the values:
E = (1/2)(1000 N/m)(0.5 m)^2 = 125 J
Therefore, the energy associated with the motion is 125 Joules.
(c) The amplitude of the oscillation is already given in the problem as 50 cm, which is equivalent to 0.5 m. So, the amplitude of the oscillation is 0.5 meters.