The general solution xt() would have the form of a linear combination of exponential functions. If the solutions move towards a line defined by a vector, the general solution would be a linear combination of exponential functions multiplied by polynomials.
In general, when solving linear homogeneous differential equations with constant coefficients, the general solution can be expressed as a linear combination of exponential functions. Each exponential function corresponds to a root of the characteristic equation.
If the solutions move towards a line defined by a vector, it means that the roots of the characteristic equation are all real and equal to a constant value, which corresponds to the slope of the line. In this case, the general solution would include terms of the form e^(rt), where r is the constant root of the characteristic equation.
To form the complete general solution, additional terms in the form of polynomials need to be included. These polynomials account for the presence of the line defined by the vector. The degree of the polynomials depends on the multiplicity of the root in the characteristic equation.
Overall, the general solution xt() in this scenario would have a combination of exponential functions multiplied by polynomials, where the exponential functions account for the movement towards the line defined by the vector, and the polynomials account for the presence of the line itself.
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a racquetball strikes a wall with a speed of 30 m/s and rebounds in the opposite direction with a speed of 1 6 m/s. the collision takes 5 0 ms. what is the average acceleration (in unit of m/s 2 ) of the ball during the collision with the wall?
The average acceleration of the racquetball during the collision with the wall is -280 m/s^2.
To find the average acceleration of the racquetball during the collision with the wall, we can use the formula:
Average acceleration = (final velocity - initial velocity) / time
Given that the racquetball strikes the wall with an initial speed of 30 m/s and rebounds with a final speed of 16 m/s, and the collision takes 50 ms (or 0.05 s), we can substitute these values into the formula:
Average acceleration = (16 m/s - 30 m/s) / 0.05 s
Simplifying this equation, we get:
Average acceleration = (-14 m/s) / 0.05 s
Dividing -14 m/s by 0.05 s gives us an average acceleration of -280 m/s^2. The negative sign indicates that the acceleration is in the opposite direction of the initial velocity, which means the ball is decelerating during the collision.
Therefore, the average acceleration of the racquetball during the collision with the wall is -280 m/s^2.
The average acceleration of the racquetball during the collision with the wall can be found using the formula:
average acceleration = (final velocity - initial velocity) / time. Given that the initial speed is 30 m/s, the final speed is 16 m/s, and the collision takes 50 ms (or 0.05 s), we can substitute these values into the formula. By subtracting the initial velocity from the final velocity and dividing by the time, we find that the average acceleration is -280 m/s^2.
The negative sign indicates that the acceleration is in the opposite direction of the initial velocity, meaning the ball is decelerating during the collision.
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A piano tuner stretches a steel piano wire with a tension of 765 N. The steel wire has a length of 0. 600m and a mass of 4. 50g.
What is the frequency f1 of the string's fundamental mode of vibration?
Express your answer numerically in hertz using three significant figures
The frequency f₁ of the string's fundamental mode of vibration is approximately 96 Hz, expressed to three significant figures.
The formula used to determine the frequency of a string's fundamental mode of vibration is given by:
f₁ = (1/2L) √(T/μ)
where:
f₁ is the frequency of the string's fundamental mode of vibration
L is the length of the string
T is the tension in the string
μ is the linear mass density of the string
Given values:
L = 0.600 m
T = 765 N
μ = 0.0075 kg/m
By substituting the values into the formula:
f₁ = (1/2L) √(T/μ)
f₁ = (1/2 × 0.600 m) √(765 N/0.0075 kg/m)
f₁ = (0.300 m) √(102000 N/m²)
f₁ = (0.300 m) (319.155)
f₁ = 95.746 Hz ≈ 96 Hz
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