Answer:
a) T = 6.49*10^-3 s
b) v = 8 m/s
c) E = 3 J
d) F = 733 N
e) F = 366.5 J
Explanation:
Given
Mass of particle, m = 94 g = 0.094 kg
Amplitude of the particle, A = 8.3 mm = 8.3*10^-3 m
Maximum acceleration of particle, a = 7.8*10^3 m/s²
the equation describing Simple Harmonic Motion is given as
x = A cos (wt +φ)
To fond the acceleration of this relationship, we would have to integrate. Twice, the first would be a Velocity, and the second acceleration that we need.
Velocity = dx/dt = -Aw sin(wt + φ)
Acceleration = d²x/dt = -Aw² cos(wt + φ)
From the question, we were given, magnitude of acceleration to be 7.8*10^3 m/s²
Aw² = 7.8*10^3
w² = 7.8*10^3 / A
w² = 7.8*10^3 / 8.3*10^-3
w² = 939759
w = √939759
w = 969
Recall, T = 2π/w, so that
T = (2 * 3.142) / 969
T = 6.49*10^-3 s
Maximum speed = Aw
Maximum speed = 8.3*10^-3 * 969
Maximum speed = 8.0 m/s
Total mechanical energy oscillator =
mgx + 1/2mx² =
1/2mv(max)² =
1/2 * 0.094 * 8² =
3 J
Maximum displacement
x = A cos(wt + φ)
For x to be maximum here, then cos(wt + φ) Must be equal to 1
Acceleration = d²x/dt² = -Aw²
And force = mass * acceleration
Force = 0.094 * 7.8*10^3
Force = 733 N
x = A cos(wt + φ), where cos(wt + φ) = 1/2
d²x/dt² = -Aw² * 1/2
d²x/dt² = 733 * 0.5
= 366.5 N
