A 10 g particle undergoes Simple Harmonic Motion with an amplitude of 2.0 mm, a maximum acceleration of magnitude 8.0 X 10^3 m/s^2, and an unknown phase constant, theta. What are a.) the period of the motion, b.) the maximum speed of the particle, and c.) the total mechanical energy of the oscillator? What is the magnitude of the force on the particle when the particle is at d.) its maximum displacement and e.) half its maximum displacement?
To find the answers to these questions, we need to use the formulae for simple harmonic motion.
a.) To find the period of the motion, we can use the formula:
T = 2π√(m/k)
where T is the period, m is the mass of the particle, and k is the spring constant.
In this case, we are given that the mass of the particle is 10 g, which is equal to 0.01 kg. However, we are not given the spring constant directly. Instead, we are given the maximum acceleration of the particle.
The maximum acceleration, a_max, is related to the spring constant by the equation:
a_max = kA
where A is the amplitude of the motion.
In this case, the maximum acceleration is 8.0 x 10^3 m/s^2 and the amplitude is 2.0 mm, which is equal to 0.002 m. Using this information, we can determine the spring constant:
k = a_max / A
= (8.0 x 10^3) / (0.002)
= 4.0 x 10^6 N/m
Now we can substitute the values into the equation for the period:
T = 2π√(m/k)
= 2π√(0.01 / 4.0 x 10^6)
≈ 0.0063 s
So, the period of the motion is approximately 0.0063 seconds.
b.) To find the maximum speed of the particle, we can use the equation:
v_max = ωA
where v_max is the maximum speed, A is the amplitude of the motion, and ω is the angular frequency.
The angular frequency can be calculated using the formula:
ω = √(k/m)
Substituting the values we found earlier:
ω = √(4.0 x 10^6 / 0.01)
= 2 x 10^3 rad/s
Now we can substitute the values into the equation for the maximum speed:
v_max = ωA
= (2 x 10^3) (0.002)
= 4 m/s
So, the maximum speed of the particle is 4 m/s.
c.) The total mechanical energy of the oscillator can be calculated using the formula:
E_total = 0.5 k A^2
Substituting the values we found earlier:
E_total = 0.5 (4.0 x 10^6) (0.002)^2
= 8 J
So, the total mechanical energy of the oscillator is 8 Joules.
d.) The force on the particle when it is at its maximum displacement can be found using Hooke's Law, which states that the force exerted by a spring is proportional to the displacement from its equilibrium position:
F = -kx
At the maximum displacement, the particle is at the amplitude of the motion, which is 2.0 mm or 0.002 m. Substituting the values we found earlier:
F = -(4.0 x 10^6) (0.002)
= -8000 N
So, the magnitude of the force on the particle at its maximum displacement is 8000 N.
e.) The force on the particle when it is at half its maximum displacement can also be found using Hooke's Law. At half the maximum displacement, the particle is at 1.0 mm or 0.001 m. Substituting the values we found earlier:
F = -(4.0 x 10^6) (0.001)
= -4000 N
So, the magnitude of the force on the particle at half its maximum displacement is 4000 N.