A horizontal spring-mass system has low friction, spring stiffness 210 N/m, and mass 0.6 kg. The system is released with an initial compression of the spring of 11 cm and an initial speed of the mass of 3 m/s.
suppose that there is energy dissipation of 0.02 J per cycle of the spring-mass system. What is the average power input in watts required to maintain a steady oscillation?
Well, well, well, looks like we have a spring-mass system that just can't stay still! Let's see if we can calculate the average power input to keep this thing oscillating.
First things first, let's figure out the total energy dissipated per second. Since we know that there is an energy dissipation of 0.02 J per cycle, and the system oscillates with a frequency of 1 cycle per second, then the total energy dissipated per second is... 0.02 J! So, that's one mystery solved.
Now, let's find the total mechanical energy of the system. The mechanical energy can be calculated as the sum of the potential energy stored in the spring and the kinetic energy of the mass. The potential energy stored in the spring can be found using Hooke's Law, which states that the potential energy is proportional to the square of the displacement. In this case, the displacement is 11 cm (or 0.11 m), and the spring stiffness is 210 N/m. So, the potential energy stored in the spring is 0.5 * 210 * (0.11)^2 J.
The kinetic energy of the mass can be calculated using the formula KE = 0.5 * mass * velocity^2. In this case, the mass is 0.6 kg, and the velocity is 3 m/s, so the kinetic energy is 0.5 * 0.6 * (3)^2 J.
Now, let's add up the potential and kinetic energies to find the total mechanical energy of the system.
By using my handy-dandy calculator, I find that the potential energy is 1.155 J and the kinetic energy is 2.7 J. So, the total mechanical energy is the sum of these two: 1.155 J + 2.7 J = 3.855 J.
Now, we can calculate the average power input required to maintain this oscillation. The average power can be found by dividing the total energy dissipated per second by the time taken for one cycle.
Since the frequency of oscillation is 1 Hz, the time taken for one cycle is 1 second. So, the average power input required is 0.02 J / 1 s = 0.02 W.
Voila! The average power input required to keep this spring-mass system oscillating is 0.02 watts. Just a tiny amount of power to unleash those springy shenanigans!
To find the average power input required to maintain a steady oscillation in the spring-mass system, we can calculate the total energy dissipated per cycle and divide it by the time taken for each cycle.
1. Calculate the total energy dissipation per cycle:
The energy dissipation per cycle is given as 0.02 J.
2. Calculate the period of oscillation:
The period (T) of an oscillating system can be calculated using the formula:
T = 2π√(m/k)
Where:
m is the mass (0.6 kg) and
k is the spring stiffness (210 N/m).
Plugging in the given values, we get:
T = 2π√(0.6/210) ≈ 0.514 s
3. Calculate the frequency of oscillation:
The frequency (f) can be calculated as the reciprocal of the period:
f = 1/T ≈ 1/0.514 ≈ 1.947 Hz
4. Calculate the average power input:
Average power (P) is given by the formula:
P = Energy dissipated per cycle / Time taken for each cycle
The time taken for each cycle is the reciprocal of the frequency:
P = 0.02 J / (1/f) = 0.02 J * f
Plugging in the value of f, we get:
P ≈ 0.02 J * 1.947 Hz ≈ 0.0389 W
Therefore, the average power input required to maintain a steady oscillation in the spring-mass system is approximately 0.0389 watts.
To find the average power input required to maintain a steady oscillation, we need to calculate the total energy dissipated per cycle and then divide it by the time taken for one cycle.
First, let's find the total energy dissipated per cycle.
The energy dissipated per cycle is given as 0.02 J.
Next, we need to calculate the time taken for one complete cycle.
The time period (T) of an oscillation can be calculated using the formula:
T = 2π√(m/k)
where m is the mass (0.6 kg) and k is the spring stiffness (210 N/m).
Let's calculate the time period:
T = 2π√(0.6/210)
T ≈ 2π√(0.0029)
T ≈ 0.341 seconds (rounded to three decimal places)
Now, we can calculate the average power input required to maintain the oscillation:
Power = Energy dissipated per cycle / Time period
Power = 0.02 J / 0.341 s
Power ≈ 0.0586 W (rounded to four decimal places)
Therefore, the average power input required to maintain a steady oscillation is approximately 0.0586 watts.