a) The resultant force on the particle is equal to - 225 · x, where x is measured in meters.
b) The particle moves with simple harmonic motion.
c) The motion has a period of approximately 1.257 seconds.
d) The particle has an approximate speed of 2.488 meters per second when it is 0.05 metres from C.
e) The equation for the position of the particle is x(t) = 0.5 · cos 25t, where t is in seconds.
How to analyze a system with a particle and two springs
In this case we have a system with a particle under periodic motion due to two reactive forces from two springs. a) The system is represented by the following formula based on Newton's laws:
∑F = - k₁ · x - k₂ · x = m · a (1)
Where:
- k₁, k₂ - Spring constants
- x - Distance of elongation, in metres.
- m - Mass of the particle, in kilograms.
- a - Net acceleration, in metres per square second.
If we know that k₁ = k₂ = 112.5 N/m, then the resultant force on the particle is equal to - 225 · x, where x is measured in meters.
b) The particle is under simple harmonic motion has a differential equation of the form:
x'' + ω² · x = 0 (2)
Where:
- x'' - Net acceleration, in metres per square second.
- x - Position of the particle, in metres.
- ω - Angular frequency, in radians.
By some algebraic handling on (1), we find the following differential equation:
x'' + [(k₁ + k₂) / m] · x = 0 (3)
Thus, the particle moves with simple harmonic motion.
c) The period of the motion (T), in seconds, is determined by T = 2π / ω. Then, the period is described by the following expression:
T = 2π · √[m / (k₁ + k₂)]
T = 2π · √(9 kg / 225 N /m)
T ≈ 1.257 s
The motion has a period of approximately 1.257 seconds.
d) The speed of the particle (v), in metres per second, can be found by the principle of energy conservation:
(1 / 2) · k · A² = (1 / 2) · k · x² + (1 / 2) · m · v² (4)
v = √[k · (A² - x²) / m]
Where:
- A - Amplitude, in metres.
- x - Particle position with respect to equilibrium position, in metres.
If we know that k = 225 N / m, A = 0.5 m, x = 0.05 m and m = 9 kg, then the speed of the particle is:
v = √[(225 N / m) · [(0.5 m)² - (0.05 m)²] / (9 kg)]
v ≈ ± 2.488 m / s
The particle has an approximate speed of 2.488 meters per second when it is 0.05 metres from C.
e) The function position for a particle under simple harmonic motion:
x(t) = A · cos ω · t (5)
If we know that A = 0.5 m and ω = 25 rad / s, then the equation for the position of the particle is:
x(t) = 0.5 · cos 25t
The equation for the position of the particle is x(t) = 0.5 · cos 25t.
To learn more on simple harmonic motion: https://brainly.com/question/17315536
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