Suppose that a 200g mass (0.20kg) is oscillating at the end of a spring upon a horizontal surface that is essentially friction-free. The spring can be both stretched and compressed and have a spring constant of 240N/m. It was originally stretched a distance of 12cm (0.12m) from its equilibrium (un-stretched) position prior to release. What is the maximum velocity that the mass will reach in its oscillation? Where in the motion is this maximum reached?
Initial kinetic energy = (1/2)kX^2
= 120*0.12^2 = 1.728 J
Set that equal to the maximum kinetic energy, and solve for Vmax.
(1/2)MVmax^2 = 1.728 J
Maximum speed is attained when the spring is unstressed (equilibrium position).
To find the maximum velocity reached by the mass in its oscillation, we can use the concept of potential and kinetic energy.
The potential energy of the mass-spring system is given by the equation:
Potential energy (PE) = (1/2) * k * x^2
where k is the spring constant and x is the displacement from equilibrium position.
In this case, the mass-spring system was originally stretched a distance of 0.12m (12cm) from its equilibrium position. Therefore, the potential energy at this point is:
PE = (1/2) * 240 N/m * (0.12m)^2
= 1.728 J
The maximum velocity is reached when all of the potential energy is converted into kinetic energy.
The kinetic energy (KE) of the mass is given by the equation:
Kinetic energy (KE) = (1/2) * m * v^2
where m is the mass and v is the velocity.
Setting the potential energy equal to the kinetic energy, we have:
PE = KE
1.728 J = (1/2) * 0.20 kg * v^2
Simplifying the equation, we get:
v^2 = (2 * 1.728 J) / 0.20 kg
= 17.28 m^2/s^2
Taking the square root of both sides, we find:
v = sqrt(17.28 m^2/s^2)
= 4.16 m/s
So, the maximum velocity reached by the mass in its oscillation is 4.16 m/s.
As for where this maximum is reached in the motion, it occurs when the mass crosses the equilibrium (un-stretched) position. In other words, the maximum velocity is reached when the mass changes direction from moving in one direction to moving in the opposite direction.
To find the maximum velocity that the mass will reach in its oscillation, we can use the conservation of mechanical energy principle.
The potential energy stored in the stretched/compressed spring is given by the formula: PE = (1/2)kx^2, where k is the spring constant and x is the displacement from the equilibrium position.
The initial potential energy of the spring when stretched 12cm (0.12m) is:
PE_initial = (1/2) (240 N/m) (0.12m)^2
When the mass reaches its maximum velocity, all the potential energy is converted into kinetic energy. Therefore, the initial potential energy is equal to the maximum kinetic energy of the mass:
PE_initial = (1/2)mv_max^2, where m is the mass and v_max is the maximum velocity.
Equating the two expressions and solving for v_max:
(1/2) (240 N/m) (0.12m)^2 = (1/2) (0.2kg) v_max^2
Simplifying the equation:
(240 N/m) (0.12m)^2 = (0.2kg) v_max^2
Now, we can solve for v_max:
v_max^2 = [(240 N/m) (0.12m)^2] / (0.2kg)
v_max^2 = (28.8 N/m) / (0.2 kg)
v_max^2 = 144 m^2/s^2
Taking the square root of both sides, we get:
v_max ≈ 12 m/s
So, the maximum velocity that the mass will reach in its oscillation is approximately 12 m/s.
Now, to determine where in the motion the maximum velocity is reached, we need to consider the nature of simple harmonic motion.
In simple harmonic motion, the maximum speed (velocity) is reached at the equilibrium position, i.e., when the mass crosses the equilibrium point. Therefore, the maximum velocity occurs when the mass is at the maximum displacement.
In this case, the maximum displacement is 0.12m, which means that the maximum velocity is reached when the mass is at 0.12m from the equilibrium (either stretched or compressed).
Please note that this explanation assumes ideal conditions and neglects any damping effects.